# If the rate of change of the acceleration of a car is 4 m/s^3. What is the velocity after 6 seconds if the initial velocity is 18 kmph.

### 2 Answers | Add Yours

`(d^2v)/(dt^2)=4m//s^3`

Taking the antiderivative we get

`(dv)/(dt)=4t + a_0`

And again to get our final answer

`v = 4*1/2 t^2 + a_0t + v_0`

Assuming a_0 = 0, and `v_0 = 18"km/hr" = 18*(1000"m/km")/(3600"s/hr") "km/hr" = 5 "m/s"`

So v = 2(6)^2+0+5"m/s"=72+5=77"m/s"=(77*1000"m/km")/(3600"s/hr")m/s=277.2 km/hr

This problem involves of the mathematical concept of integration.

For an object that is moving, the rate of change of its velocity is its acceleration. The rate of change of acceleration is given as 4 m/s^3.

If the velocity is denoted by v, acceleration is given as `a = (dv)/(dt)`

Here, `(da)/(dt) = 4`

=> `(d^2v)/dt^2 = 4`

Integrating this twice gives the velocity in terms of the time.

`int 4 dt = 4t + C`

`int 4t + C dt`

=> 2t^2 + C1*t + C2

It is not given at what time the velocity is 18 km/h. Also, there are two variables C1 and C2 of which only one can be estimated. If it is assumed that the initial velocity is the velocity at t = 0 s and C1 = 0

=> 2*0 + C2 = 5 m/s

Substituting C2 = 5 and C1 = 0 in the expression for velocity: V = 2*t^2 + 5 m/s

After 6 seconds, the velocity is 2*36 + 5 = 77 m/s = 77*3600/1000 km/h = 277.2

**The velocity of the car after 6 seconds is 277.2 km/h**