The engine of a car provides a forward force of \(1240\ \mathrm{N}\) and the total resistive force on the car is \(800\mathrm{N}\). If the mass of the car i...

Answer Details

Acceleration[a] = [forward−resistive]forcemass

a = 1240−8001220 $\frac{1240-800}{1220}$ → 440−8001220 $\frac{440-800}{1220}$

a = 0.36m/s2 ${}^2$,  v = 4m/s, s = ?

Second equation of motion:

s is distance
u is initial velocity
v is final velocity
a is acceleration

v2 ${}^2$ = u2 ${}^2$ + 2as

42 ${}^2$ = 02 ${}^2$ + 2 * 0.36 * s

16 =  0.72s 

22.2m = s