Question

Austin's truck has a mass of 2000 kg when traveling at 22.0 m/s, it brakes to a stop in 4.0 s. show that the magnitude of the braking force acting on the truck is 11,000 n

Answer 1

Since F=m•a, you want to show that a = -5.5

Answer 2

Since the truck was in motion before the brakes were applied, it will decelerate within the 4s before coming to a stop. Hence the acceleration is
`-a\:ms^(-2)`.

When the truck comes to a stop, It will have a final velocity of
`v=0\:ms^(-1)`.

Also the initial velocity is 22.0 m/s. This means,
`u=22.0\:ms^(-1)` and time,
`t=4s`.

We can use the relation,

`v=u+at`

to determine the acceleration of the truck.

Let us now plug in all the values to obtain,

`0=22.0+(-a)(4)`

`\Rightarrow -22.0=-4a`

`\Rightarrow (-22.0)/(-4)=a`

`\Rightarrow a=5.5 ms^(-2)`

Using the relation,

`F=ma`

the magnitude of the braking force is

`F=2000 * 5.5 N=11000N`