# A car accelerated from rest uniformly at 2.40 m/s2 for 15.0 s, then moves with uniform velocity for 200s, and finally decelerates uniformly at 3.60m/s2.The car comes to a stop after decelerating...

A car accelerated from rest uniformly at 2.40 m/s2 for 15.0 s, then moves with uniform velocity for 200s, and finally decelerates uniformly at 3.60m/s2.

The car comes to a stop after decelerating uniformly at 3.60m/s2.

How long will the car be in motion , how far will it travel, and what will be its average speed?

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### 2 Answers

For the acceleration, the average was 19.2 m/s2, found by adding all the speeds together over the 15 second time frame, then dividing by 15. For the deceleration, the average was 16.2 m/s2, found again by adding all the speeds together per the 10 second time frame for deceleration, then dividing by 10. The total amount of time the car was in motion is easiest, found by adding the 15 seconds of acceleration to the 200 seconds of constant speed, to the 10 seconds of deceleration. This gave a total of 225 seconds for total motion of the car. To calculate the total distance, for the acceleration, the car is covering 2.4 meters per second/per second, for a total of 288 meters. During the constant speed, it covered 36 meters/second for 200 seconds. for a total of 7200 meters. For the deceleration, the car covered 162 meters. If you add the 288 + 7200 + 162, you get a total distance of 7650 meters. If you take that distance and divide it by the total time, you get the average speed: 7650 meters/225 seconds = 34 meters per second.

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### User Comments

Initial velocity u=0

Final velocity v=?

Acceleration a=2.4 m/s2

Time t=15 s

We know,

a=(v-u)/t

2.4=(v-0)/15

36=v-0

v=36 m/s ………………………….(i)

let, distance travelled s=?

we know,

s=ut+1/2at2

s=540 m ………………………….(ii)

again,

speed=36 m/s

time=200 s

therefore, distance travelled=(36x200) m = 7200 m ………………………….(iii)

again,

initial velocity u=36 m/s

final velocity v=0

retardation a’=3.6 m/s2

time t=?

we know,

a’=(v-u)/t

3.6t=-36

t=-10 = 10 s ……………………………(iv)

again,

s=ut+1/2at2

s=(360+180) m = 540 m ……………………………….(v)

therefore,

the car will be in motion for (15+200+10)s = 225 s

total distance travelled = (540+7200+540)m = 8280 m

avg. speed = (8280/225)m/s = 36.8 m/s