Force and Laws of Motion — Question 4

Given,

Mass of the car (m) = 1200 kg

Initial velocity (u) = 90 kmh^-1

Convert kmh^-1 to ms^-1 : multiply by $\dfrac{5}{18}$

90 x $\dfrac{5}{18}$ = 25 ms^-1

Final velocity (v) = 18 kmh^-1

Convert kmh^-1 to ms^-1 : multiply by $\dfrac{5}{18}$

18 x $\dfrac{5}{18}$ = 5 ms^-1

Time period (t) = 4 sec

As per the first equation of motion,

a = $\dfrac{\text{v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{5-25}{4}$ = -5ms^-2

Hence, acceleration of the car is -5 ms^-2

Negative sign shows retardation.

Initial momentum of the car = m × u = 1200 × 25 = 30,000 kgms^-1

Final momentum of the car = m × v = 1200 x 5 = 6,000 kgms^-1

∴ Change in momentum = mv - mu = 6,000 - 30,000 = -24,000 kgms^-1

Hence, change in momentum = 24,000 kgms^-1

We know, Force = mass x acceleration

Substituting we get,

F = 1200 × -5 = -6000 N

Hence, the magnitude of force required to slow down the vehicle = 6000 N