A car is originally moving at 19.9 m/s in the positive x with a constant acceler

Question

A car is originally moving at 19.9 m/s in the positive x with a constant acceleration of 8.2 m/s2 in the positive x direction for 12.6 seconds. The car then hits its it brakes and decelerates at a constant rate until it stops after additional 2.9 seconds (the brakes are applied for 2.9 seconds). After it stops, what is the average speed of the car for the entire trip until it stops in m/s?

An object is originally moving at 3.5 m/s in the negative x direction with a constant acceleration of 2.4 m/s2 in the negative x direction for 5.1 seconds. Then (immediately) a constant acceleration is applied to in the positive x direction. After some time, the average velocity for the entire trip is 0 m/s., and at that time, the object is moving at 61.2 m/s in the positive x direction. How long was the acceleration applied in the positive x direction in seconds?

Explanation / Answer

For average speed we will need total distance,

during acceleration, d1 = ut + at^2 /2

d1 = 19.9 x 12.6 + 8.2 x 12.6^2 /2 = 901.66 m

speed at the end of 12.6 sec v = 19.9 + 8.2 x12.6 =123.22 m/s

during decceleration,

v = u + at

0 = 123.22 + a*2.9

a = -42.49 m/s

using v^2 - u^2 = 2ad2

0 - 123.22^2 =2 * 42.49 * d2

d2 = 178.67 m

dtotal = d1 + d2 = 1080.33 m

time = 12.6 + 2.9 = 15.5 sec

v = d /time = 69.70 m/s

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