a bus with a maximum speed of 20m/s takes 21sec to travel 270m from stop to stop. Its acceleration is twice as great as its deceleration. Find: A, the acceleration B, the distance traveled at maximum speed​

Answer:

We can use the kinematic equations to solve this problem.

Let a be the acceleration and d be the distance traveled at maximum speed.

First, we can use the equation:

d = vt + (1/2)at^2

where v is the maximum speed, t is the time traveled at maximum speed, and a is the acceleration.

We know that the bus takes 21 seconds to travel 270 meters, so the time traveled at maximum speed is:

21 - 2a = t

We also know that the acceleration is twice as great as the deceleration, so we can write:

a = 2d

Then, we can substitute these expressions into the first equation:

d = (20)(21 - 2a) + (1/2)(2d)(21 - 2a)^2

Simplifying and solving for d, we get:

d = 210 - 5.25a + 0.25a^2

To find the acceleration, we can use the fact that the maximum speed is reached at the midpoint of the trip, so the distance traveled at maximum speed is half the total distance:

d = 1/2(270)

Solving for d, we get:

d = 135

Substituting this value into the equation for d, we get:

135 = 210 - 5.25a + 0.25a^2

Simplifying and solving for a, we get:

a = 8 m/s^2

Finally, we can use the equation:

d = vt

to find the distance traveled at maximum speed:

d = (20)(21 - 2a)

Substituting the value of a, we get:

d ≈ 188.4 m

Therefore, the acceleration of the bus is 8 m/s^2 and the distance traveled at maximum speed is approximately 188.4 meters.

What is the average acceleration of an
automobile moving from 0 to 60 m/s in 3.0
seconds?
a. 90 m/s²
b. -20 m/s²
c. -90 m/s²
d. 20 m/s²