Question

If a car with velocity `100 "km"/"hr"` slams on the brakes with constant deceleration of `10 "m"/("s"^2)`, how far does the car travel before coming to a complete stop?

Using integrals as general (and particular) solutions to a first order, linear, differential equation.

Answer 1

`approx2.78sec` or `100/36sec`

Explanation:

Using `suvat` equations,
`v=u+at`
`0=(100*1000)/(60*60)+(-10)t`
`10t=100000/3600`
`t=100/36secapprox2.78sec`

Answer 2

`t=2.78" s"`

Explanation:

Deceleration is just negative acceleration.

Acceleration is `a(t)=-10"m"/("s"^2)`

Velocity is the antiderivative of acceleration.

`v(t)=int-10dt`

`v(t)=-10t+C_1`

Because the initial velocity at `t=0` was `100"km"/"h"`, but acceleration is in meters per second, per second, we need to convert this velocity to meters per second.

`100cancel("km")/cancel("h")xx(1000" m")/(1cancel(" km"))xx(1cancel(" h"))/(3600" s")=27.8 "m"/"s"`

So, at time `t=0`, we have

`v(0)=-10(0)+C_1=27.8`

`=> C_1=27.8`

`v(t)=-10t+27.8`

We want to know the time when the car stops, which means when the velocity is zero.

`v(t)=-10t+27.8=0`

`27.8=10t`

`t=2.78" s"`