# The car is at rest at t = 0. It has a velocity given by v = 20t- 0.5t^2 km/h, where t is in seconds. What is the car’s maximum acceleration and what is the distance traveled in 15 s.

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### 1 Answer

This question has been moved to the Math section as the use of mathematical symbols are required to arrive the answer.

The car is at rest at t = 0. It has a velocity given by `v = 20*t - 0.5*t^2` km/h, where t is in seconds.

First make the units the same in the expression for v,

v = `20*t - 0.5*t^2` km/h

=> `(20*t - 0.5*t^2)*(1000/3600)` m/s

a = `(dv)/(dt)` = `(1000/3600)*(20 - t)` m/s^2

To find the maximum acceleration solve `(da)/(dt) = 0`

=> `-t = 0`

=> `t = 0`

The acceleration is maximum when the car starts and is equal to `50/9 ` m/s^2

The distance traveled in 15 s is given by the integral:

`int_(0)^15 (20*t - 0.5*t^2)*(1000/3600)`

=> `(1000/3600)*(20t^2/2 - 0.5*t^3/3)` between t = 0 and t = 15

=> `(1000/3600)*(20*15^2/2 - 0.5*15^3/3)`

=> 468.75 m

**The maximum acceleration of the car is `50/9` m/s^2 and the distance traveled in the first 15 s is 468.75 m**

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