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.

1 Answer | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

Sources:

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question