Question

What average net force is required to bring a 1640kg car to rest from a speed of 75km/h within a distance of 90m?

Answer 1

`3954.04 N`

Explanation:

newton's 2nd law: `F = ma`

force = mass * acceleration

`v^2 = u^2 + 2as`, where

• `u` is initial velocity

• `v` is final velocity

• `a` is acceleration

• `s` is displacement (in this case, distance is used).

the SI units for speed are `m//s`.

to convert `75 km//h` to `m//s:`

`75 km//h = (75/3600)km//s`

`(75/3600)km//s = (75/3.6) m//s`

`75/3.6 = 125/6`

`75 km//h = 125/6 m//s`

here, the initial velocity `u` is `125/6 m//s`.

the final velocity `v` is `0` `m//s`, since the car is being brought to rest.

the displacement `s` is `90m`.

`v^2 = u^2 + 2as`

`0 = (125/6)^2 + 180a`

subtract `(125/6)^2` from both sides:

`180a = -(125/6)^2`

`180a = -15625/36`

`a = -15625/(36 * 180) = -15625/6480`

`a = -2.411` (3s.f.)

acceleration `= -2.411 m//s^2`
deceleration `= 2.411 m//s^2`

`F = ma`

this means that the net force can be found by multiplying the mass of the car by its acceleration.

the force in question is the net force, meaning that the positive value of `a` can be used.

`F = 1640 kg * 2.411 m//s^2`

`= 3954.04 N`