During a relay race, runner A runs a certain distance due north and then hands off the baton to runner B, who runs for the same distance in a direction south of east. The two displacement vectors A and B can be added together to give a resultant vector R. Which drawing correctly shows the resultant vector?

Answer:

Velocity and Time

Explanation:

I googled it-

Answer:

The answer is below

Explanation:

Given that:

y = transverse displacement = 4.2 cm = 0.042 m at x = 0 and t = 0.

Speed = v = 89 m/s, maximum transverse speed of the string particle = [tex]u_m[/tex] = 16 m/s.

ω = [tex]u_m[/tex] / [tex]y_m[/tex] = 16 / 0.42 = 380.95 rad/s

a) Frequency = ω/2π = 380.95 / 2π = 60.63 Hz

b) Wavelength (λ) = speed / frequency

λ = v / f = 89/63.66= 1.468 m

c) Using the wave equation:

[tex]y=y_msin(kx \pm wt \pm \phi)\\y=0.042,t=0,x=0\\\\Hence\\y_m=0.042\ m[/tex]

d) Wave number k is given by:

k = 2π / λ = 2π / 1.468 = 4.28 rad/s

e) The angular velocity is given by:

ω = [tex]u_m[/tex] / [tex]y_m[/tex] = 16 / 0.42 = 380.95 rad/s

f)  Using the wave equation:

[tex]y=y_msin(kx \pm wt \pm \phi)\\\\y=0.042,t=0,x=0,y_m=0.042\\\\Hence\\0.042=0.042sin(4.28(0)\pm 380.95(0)\pm \phi)\\\\sin\phi=1\\\\\phi=\frac{\pi}{2} \\\\y=0.042sin(4.28x\pm 380.95t\pm \frac{\pi}{2})[/tex]

g) Since the wave is in the positive x direction, hence ω is negative

Explanation:

Weight of an object is shows the force of gravity acting on it. It is calculated mass times acceleration due to gravity.

lb is the unit of avoirdupois weight. It means that 12 lb shows weight.

gram, kg is the unit of mass. It means 0.34 g and 120 kg shows mass of an object.

m and cm are the units of length. It means 0.34 m and 411 cm shows the length of the object.

N is the unit of weight.

Hence, 12.0 lb and 1600 kN are acceptable representations of weight.

Complete Question

A man on a road trip drives a car at different constant speeds over several legs of the trip. He drives for 55.0 min at 60.0 km/h, 18.0 min at 80.0 km/h, and 60.0 min at 60.0 km/h and spends

25.0 min eating lunch and buying gas

What is the total distance traveled over the entire trip (in km)

Answer:

The value is [tex]D = 139.02 \ km[/tex]

Explanation:

From the question we are told that

For [tex] t_1 = 55 min = \frac{55}{60} = 0.917 \ h[/tex] the speed is [tex]v_1 = 60 \ km/h[/tex]

For [tex] t_2 = 18 min = \frac{18}{60} = 0.3 \ h[/tex] the speed is [tex]v_2 = 80 \ km/h[/tex]

For [tex] t_3 = 60 min = \frac{60}{60} = 1 \ h[/tex] the speed is [tex]v_3 = 60\ km/h[/tex]

The time taken to have lunch is [tex]t _l = 25 \ min = \frac{25}{60} = 0.42 \ h[/tex]

Generally the total distance traveled over the entire trip (in km) is mathematically represented as

[tex]D = t_1 * v_1 + t_2 * v_2 + t_3 * v_3[/tex]

=>   [tex]D = 0.917  *  60 +  0.3 * 80 +  1  * 60[/tex]

=>   [tex]D = 55.02  + 24 +   60[/tex]

=>   [tex]D = 139.02 \ km[/tex]

The magnetic field is just a unit vector that explains the electromagnetic influence on electric currents, current flow, and magnetic fluids.

Current density [tex]J = ar^2[/tex]

value of the constant is=  [tex]28.5 \ \frac{A}{mm^4}[/tex]

radius = 5.20 mm

magnetic permeability [tex]\mu = 4\pi \times 10^7 \ \frac{N}{A^2}[/tex]

calculating the area element for the straight circular conduction rod
:

[tex]d_A=2\pi r dr[/tex]

Calculating the current, which is carried in the rod

[tex]dI = \int dA \vec{J}[/tex]

Calculating the above equation with the limit value that is 0 to r.

[tex]I^{1}=\int_{0}^{r} ar^2 \times 2 \pi \cdot r d_r[/tex]

    [tex]=2\pi a\int_{0}^{r} r^3 d_r \\\\=\frac{\pi ar^4}{2}[/tex]

The calculated current value which is carried by the rod is [tex]\boxed{\frac{\pi ar^4}{2}}[/tex]

In option (a)

calculating the magnitude of the magnetic field at the point [tex]r_1= \frac{R}{2}[/tex]

[tex]\to B=\frac{\mu_{0} I^{1}}{2 \pi r}\\\\\to B \times 2\pi r= \mu_{0} (\frac{\pi a}{2}r^4)\\\\\to B \times 2\pi = \mu_{0} (\frac{\pi a}{2}r^3)\\\\\to B= \frac{\mu_{0}}{4}r^3 a\\\\[/tex]

Substituting the above value:[tex]\frac{R}{2} \ for \ r[/tex]

[tex]B= \frac{\mu_{0}}{4}(\frac{R}{2})^3 a[/tex]

[tex]B= \frac{4 \pi \times 10^{-7}}{4}(\frac{5.2 \times 10^{-3}}{2})^3 \frac{28.5}{10^{-12}}[/tex]

   [tex]= \frac{4 \pi \times 10^{-7}}{4} \times \frac{140.608 \times 10^{-9}}{8} \times \frac{28.5}{10^{-12}}\\\\= \frac{ 3.14 \times 10^{-7}}{1} \times \frac{140.608 \times 10^{-9}}{8} \times \frac{28.5}{10^{-12}}\\\\=\frac{1572.87624 \times 10^{-16}}{ 10^{-12}}\\\\=0.157 \ \ T[/tex]

Thus, the magnitude of the magnetic field at the point [tex]r_1 =\frac{R}{2}[/tex] is [tex]\boxed{0.157 \ T}[/tex]

In option (b)
In this, we calculate the magnitude of the magnetic field at the point [tex]r_2= 2R[/tex]

[tex]\to B=\frac{\mu_{0} I^{1}}{2 \pi r} \\\\\to B \times 2\pi r= \mu_{0} (\frac{\pi a}{2}R^4)\\\\\to B \times 2\pi 2R = \mu_{0} (\frac{\pi a}{2}R^3)\\\\\to B= \frac{\mu_{0}}{8}R^3 a\\\\[/tex]

Substituting the values  

[tex]B= \frac{4 \pi \times 10^{-7}}{8}(5.2 \times 10^{-3})^3(\frac{28.5}{10^{-12}})[/tex]

   [tex]= \frac{4 \times 3.14 \times 10^{-7}}{8} \times (5.2)^3 \times (10^{-3})^3 \times\frac{28.5}{10^{-12}}\\\\= \frac{ 3.14 \times 10^{-7}}{2} \times 140.608 \times 10^{-9} \times\frac{28.5}{10^{-12}}\\\\= 6291.50495 \times 10^{-4}\\\\= 0.629 \ \ or \ \ 0.63\\[/tex]

Thus, the magnitude of the magnetic field at the point [tex]r_2 = 2R[/tex] is [tex]\boxed{0.63 \ \ T}[/tex]

Find out more about Magnetic fields here:

brainly.com/question/14848188

Explanation:

1. Given:

v₀ = 0 m/s

a = 2.0 m/s²

t = 1.2 s

Find: v

v = at + v₀

v = (2.0 m/s²) (1.2 s) + 0 m/s

v = 2.4 m/s

2. Given:

v₀ = 0 m/s

v = 9.0 m/s

a = 2.2 m/s²

Find: t

v = at + v₀

9.0 m/s = (2.2 m/s²) t + 0 m/s

t ≈ 4.1 s

Answer:

Explanation:

For two vectors to be orthogonal (perpendicular) the product of both vectors must be zero. Given the vectors D=(2.0i-4.0j+k)N and G=(3.0i+4.0j+10k)N, to show that they are orthogonal, we will take their dot product as shown;

D.G = (2.0i-4.0j+k).(3.0i+4.0j+10k)

Note that i.1 = j.j = k.k = 1 and dot product of different component is zero.

D.G = 2.0(3.0) (i.i) + (-4.0)(4.0)j.j + 1(10)k.k

D.G = 6.0(1) -(16)(1)+10(1)

D.G = 6-16+10

D.G = -10+10

D.G = 0

Since the dot product of the two vectors is zero, this shows that force vector D is orthogonal to force vector G.

Answer:

[tex](a) 189.23 N[/tex], [tex](b) 47.31 N[/tex] and [tex](c) 141.92 N[/tex].

Explanation:

Three balls are shown in figure having charge [tex]q=1.45 \mu C[/tex]. The middle ball, [tex]B[/tex], is positively charged having charge [tex]+q[/tex], and the remaining two outside balls, [tex]A[/tex] and [tex]C[/tex], are negatively charged having charged [tex]-q[/tex] as shown.

[tex]AC=20 cm[/tex] and [tex]AB=BC=10[/tex] cm as B is the mid-point of AC.

Let [tex]d_1=AC=20\times 10^{-3}m[/tex] and [tex]d_2=AB=BC=10\times 10^{-3}m[/tex]

From Coulomb's law, the magnitude of the force, [tex]F[/tex], between two point charges having magnitudes [tex]q_1 \& q_2[/tex], separated by distance, [tex]d[/tex], is

[tex]F=\frac {1}{4\pi\epsilon_0}\frac {q_1q_2}{d^2}\;\cdots (i)[/tex]

where, [tex]\epsilon_0[/tex] is the permittivity of free space and

[tex]\frac {1}{4\pi\epsilon_0}=9\times 10^9[/tex] in SI units.

This force is repulsive for the same nature of charges and attractive for the different nature of charges.

Now, Using equations(i),

(a) The magnitude of attraction force between balls A and B is

[tex]F_{AB}=F_{BC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_2)^2}[/tex]

[tex]\Rightarrow F_{AB}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(10\times 10^{-3}\right)^2}[/tex]

[tex]\Rightarrow F_{AB}=189.23 N[/tex]

(a) The magnitude of the repulsive force between balls A and C is

[tex]F_{AC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_1)^2}[/tex]

[tex]\Rightarrow F_{AC}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(20\times 10^{-3}\right)^2}[/tex]

[tex]\Rightarrow F_{AC}=47.31 N[/tex]

(c) The magnitude of the net force, [tex]F_{net}[/tex], on the outside of the ball is,

[tex]F_{net}=189.23-47.31 N[/tex]

[tex]\Rightarrow F_{net}=141.92 N[/tex]

Answer:

0.59 seconds

Explanation:

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = [tex]v_{oy}[/tex] / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

(a) The horizontal velocity in the projectile motion is always constant.it is the horizontal component of velocity by which the object is thrown. The expression for the football horizontal velocity will be [tex]H = \frac{u^{2} sin^{2}\theta}{2g}[/tex].

(b) The amount of time it takes for the body to project and land is the time of flight.

(c) The horizontal distance traveled by the ball is defined by the ball is called the range of the ball. The horizontal distance traveled will be 16.7 m.

The horizontal distance is covered by the body when the body is thrown at some angle is known as the range of the projectile. It is given by the formula

(a) Let the velocity at which the ball passes will be v. Resolve the velocity components into two components one is the horizontal component

[tex]\rm{u_x = ucos\theta}[/tex]

[tex]\rm{u_y = usin\theta}[/tex]

let the x distance is traveled in the horizontal direction so,

[tex]y = u_y \times t[/tex]

[tex]\rm{H = u_y t+\frac{1}{2} gt^2}[/tex]

[tex]\rm{H = usin\theta (\frac{usin\theta}{g}) +\frac{1}{2} g(\frac{usin\theta}{g}) ^2}[/tex]

[tex]H = \frac{u^{2} sin^{2}\theta}{g} -\frac{u^{2}sin^{2}\theta }{2g}[/tex]

[tex]H =\frac{u^{2}sin^{2}\theta }{2g}[/tex]

This is the required relation for the horizontal velocity.

(b)

Newton's equation of motion

[tex]\rm{v = u +gt}[/tex]

[tex]\rm{v_y = u_y +gt}[/tex]

[tex]\rm{v_y = 0}[/tex]

[tex]{u_y = usin\theta}[/tex]

[tex]\rm{usin\theta = gt}[/tex]

[tex]t= \rm{ \frac{usin\theta}{g} }[/tex]

These are the required relation for time t.

(c)

The range of the projectile is given as

[tex]R =\frac{u^{2}sin{2}\theta }{g}[/tex]

[tex]R =\frac{(13.5)^{2}sin64^0 }{9.81}[/tex]

R = 16.7 m

So the horizontal distance covered will be 16.7 m.

To learn more about the range of projectile refer to the link ;

https://brainly.com/question/139913

Answer:

a)   t = 57.14 s

b)    x = 27711.4 m

Explanation:

This is a missile throwing exercise

a) They ask us for the time to the maximum reach, this corresponds to when it reaches the ground y = 0, let's use

          y =  [tex]v_{oy}[/tex] t - ½ g t²

Let's use trigonometry to find the vertical initial velocity

         sin θ = v_{oy} / v₀

         v_{oy} = v₀ sin θ

we substitute

         y = v₀ sin θ t - ½ g t²

since the height is zero

          0 = t (v₀ sin θ - ½ g t)

This equation has two solutions

*  t = 0 which corresponds to the moment of launch

*

         v₀ sin 30 - ½ g t = 0

        t = v₀ sin 30    2/g

let's calculate

         t = 560 sin 30  2 / 9.8

         t = 57.14 s

b) with this time we can calculate the distance traveled

          x = v₀ₓ t

let's use trigonometry for velocity

          cos θ = v₀ₓ / v₀

          v₀ₓ = v₀ cos 30

we substitute

         x = v₀ cos 30     t

let's calculate

         x = 560 cos 30 57.14

         x = 27711.4 m

Answer:

2.15m/26.1s is = average speed

0.08 m/s

Answer:

h = 31.9 m

Explanation:

Since, the ball took 5.1 s in the air. Hence, the time to reach maximum height will be equal to the half of this value:

t = 5.1 s /2 = 2.55 s

Now, we use 1st equation of motion between the time of throwing and the time of reaching maximum height:

Vf = Vi + gt

where,

Vf = Final Velocity = 0 m/s (since, ball momentarily stops at highest point)

Vi = Initial Velocity = ?

g = - 9.8 m/s² (negative sign for upward motion)

Therefore,

0 m/s = Vi + (-9.8 m/s²)(2.55 s)

Vi = 24.99 m/s

Now, we use second equation of motion for height (h):

h = Vi t + (0.5)gt²

h = (24.99 m/s)(2.55 s) + (0.5)(-9.8 m/s²)(2.55 s)²

h = 63.7 m - 31.8 m

h = 31.9 m

Answer:

its right the way it is

Explanation:

if there is a multiple choice then pick 20 v

Answer:

The mass in grams of a red blood cell is about 7.85 ×  10⁻¹¹ grams

Explanation:

To find the mass in grams of a red blood cell,

From,

[tex]Density = \frac{Mass}{Volume}[/tex]

Then,

[tex]Mass = Density \times Volume[/tex]

From the question,

Density of a red blood cell is similar to that of water

Density of water = 1 g/mL = 1 g/ cm³

Then, Density of a red blood cell = 1 g/cm³

Now, we will find the volume a red blood cell.

From the question,  

Since the shape is like that of a thick disc, we can determine the volume by using the formula for volume of a cylinder.

Hence,

Volume of a red blood cell = [tex]\pi r^{2}h[/tex]

Where [tex]\pi[/tex] Is a constant (Take [tex]\pi[/tex] = 3.14)

[tex]r[/tex] is the radius

and [tex]h[/tex] is the thickness

Diameter of a red blood cell = 10 micrometers

Then, radius of a red blood cell = 10/2 micrometers = 5 micrometers

[tex]r[/tex] = 5 micrometers = 5 × 10⁻⁶ meters

and [tex]h[/tex] = 1 micrometer = 1 × 10⁻⁶ meters

Hence,

Volume of a red blood cell = 3.14 × (5 × 10⁻⁶)² × 1 × 10⁻⁶

∴ Volume of a red blood cell = 7.85 × 10⁻¹⁷ cubic meter (m³)

Convert this to cubic centimeter

(NOTE: 1 cubic meter = 1000000 cubic centimeter)

Hence,

Volume of a red blood cell = 7.85 ×  10⁻¹¹ cubic centimeter (cm³)

Now, for the mass

[tex]Mass = Density \times Volume[/tex]

Density of a red blood cell = 1 g/cm³

Volume of a red blood cell = 7.85 ×  10⁻¹¹ cubic centimeter (cm³)

Then,

Mass = 1 g/cm³ ×  7.85 ×  10⁻¹¹ cm³

Mass = 7.85 ×  10⁻¹¹ g

Hence, the mass in grams of a red blood cell is about 7.85 ×  10⁻¹¹ grams

Answer:

Choice A. Without energy storage, the total work output of a machine will not exceed the total work input.

Explanation:

Work is the product of force and the distance travelled in the direction of that force.

Indeed, simple machines can alter the size or direction of a force. However, their work output can't exceed the work input to that machine (unless the machine stores some energy- by bending, for example.) Because of frictions and other forms of energy loss, the useful work output of a real machine can even be smaller than the work input to that machine.

The force output of some machines can certainly exceed their force input. However, at the same time, the input part of those machine needs to move a  distance that is (more than) proportionally longer compared to the output part of the machine. As a result, the useful work output would be equal to or even smaller than the total work input.

A fixed pulley is an example of a machine that applies a force in a direction different from the direction of the force applied to the machine. Pulling the string down will lift up the weight on the other side. A seesaw (a first-class lever) is another real-world example of this kind of machines.

Tongs are examples of machines where: the distance traveled by the output exceeds that of the input. However, that comes at a price: delivering the same amount of force with a pair of tongs requires more force than without a pair of tongs. That corresponds to choice D: the work output of some machines can be less than the force input to these machines.

At the same time, some extra force input might be required to overcome the friction between parts of the machine. That's another reason why the machine applies less force than the input even if it was designed to apply the same amount of force.

Answer:

A.) Do a greater amount of work than the amount of work done on the machine.

Explanation:

I took the quiz on Edge!

Hope this helped!

Answer:

Explanation:

When a body is held against a vertical wall , to keep them in balanced position , normal force is applied on their surface . this force creates normal reaction which acts against the normal force and it is equal to the normal force as per newton's third law . Ultimately friction force is created which is proportional to normal force and it acts in vertically upward direction . It prevents the body from falling down .

Hence normal force = reaction force .

From second law also net force is zero , so if  normal force is N and reaction force is R

R - N = mass x acceleration = mass x 0 = 0

R = N .

Ranking normal force from  highest  to smallest

150 N , 130 N , 120 N

B )

Frictional force is equal to the weight of the body because the body is held at rest .

Ranking of frictional force form largest to smallest

7 kg , 5 kg , 3 kg , 1 kg .

Here frictional force is irrespective of the normal force acting on the body  because frictional force adjusts itself so that it becomes equal to weight in all cases here because it always balances the weight of the body .

Based on the magnitude of normal force acting on them, the ranking from largest to smallest will be:

Based on the masses, the frictional force acting on the boxes will be such that the ranking from largest to smallest will be:

The normal forces acting on the box will depend on the force acting on the box. This means that the larger the force acting on the box, the larger the normal force.

This is why the box with a force of 150 N will be have the highest normal force acting on it.

The frictional force acting on the boxes will depend on the mass. Boxes will lager masses will have more frictional force acting on them.

The boxes of 7kg will therefore have the largest frictional forces acting on them.

Find out more on frictional forces at https://brainly.com/question/24386803.

Answer: Hypothesis

Explanation:

For each experiment involving nanotechnology, Gerry  and Lena will test a hypothesis.

Gerry and Lana will use the Scientific method for their nanotechnology research and a key stage in the scientific method is to come up with a Hypothesis.

A hypothesis gives a researcher direction because it is a theory that they formulate that is meant to explain the phenomenon that they are researching.

They will then test this theory to either disprove or approve it and in so doing will be able to come up with conclusions to the research. Gerry and Lana will therefore have to test a hypothesis for each experiment in order to continue with their research.

Answer:

[tex]a=-1.5\ m/s^2[/tex]

Explanation:

Given that,

Initial velocity, u = 30 m/s

Final velocity, v = 0

Time, t = 20 s

We need to find the acceleration of the car. The rate of change of velocity is equal to acceleration. So,

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{-30}{20}\\\\a=-1.5\ m/s^2[/tex]

So, the magnitude of the acceleration of the car is [tex]1.5\ m/s^2[/tex] and it is decelerating.

Answer:

when this charged glass rod is brought towards the metal ball it will acquire a charge opposite to that of charge body brought close to it without touching it but it will acquire the same charge if the charged object touches it

Answer:

B.The glass and the paper have different charges.

Explanation:

Answer: las respuestas correctas son b, c y  d.

indican cantidad

Physical properties of matter tells us what a substance is or what a substance can do when it is not undergoing a chemical change.

These properties are observable with our senses or instruments or some pieces of apparatus.

Some of these properties are :

Color

Odor

hardness

texture

boiling point

Melting point

density

viscosity

They differ from chemical properties which are shown by the reaction of a substance with another.

Answer:

Definition - The friction that occurs when an object rolls across a rolling friction is easier to overcome than sliding friction.

Ex - Anything with wheels (cars, bicycle,etc) or ball rolling.

Explanation:

I think this answer will help .... if so please follow me.

Answer:

D

Explanation:i think but dont get mad if im wrong

Answer:

. the point charge is moved off center (but still inside the sphere) the point charge is moved to allocation just outside the sphere.

Explanation:

Because

Gaussian surface is given as

ϕ= qclose/E0

​So Q enclosed charge will be same if w the charge inside the Gaussian surface is moved, so flux will not change.

​

Answer:

-10.267

Explanation:

Initial speed > 35 miles per hour

Final speed > 0 miles per hour

Time > 5 seconds

you have to divide the velocity by the time which should be 35 / 5 = 7

so the answer should be 7 but I'm not sure

Explanation:

The force on an object due to it's mass and gravity.

Answer:53.03

Explanation:

Answer:

[tex]v^2 = v_{ox}^2 + 2a(x - o_x)[/tex]

Explanation:

The options are not well presented; However, the questions can still be solved

Given

[tex]Initial\ Velocity= V_{ox}[/tex]

[tex]Initial\ Displacement = X_{o}[/tex]

[tex]Time = t[/tex]

Required

Determine the final displacement

This question will be answered using the following equation of motion

[tex]v^2 = u^2 + 2as[/tex]

Where s represent the total distance

s is the distance between the initial and final displacements and is calculated as thus;

[tex]s = x - o_x[/tex]

Where x represents the final displacement

Substitute [tex]V_{ox}[/tex] for [tex]u[/tex] ---- The initial velocity

[tex]v^2 = v_{ox}^2 + 2as[/tex]

Substitute [tex]x - o_x[/tex] for s

[tex]v^2 = v_{ox}^2 + 2a(x - o_x)[/tex]

Hence, the above equation can be used to determine the final displacement by solving for x