Question

A 1000 kg vehicle moving with a speed of 20m/s is brought to rest in a distance of 50 meters .Find acceleration .Calculate the unbalanced force acting on the vehicle

Answer 1

Mass of vehicle = 1000 kg

Initial speed (u) = 20 m/s

Distance traveled (s) = 50 meters

Final speed (v) = 0 m/s

Let the acceleration be a.

Now, using third equation of motion:

$v^2 = u^2 + 2aS$

$0^2 = 20^2 + 2 \times a \times 50$

$a = - \frac{400}{100}$

a = - 4 m/s^2 (Negative means that it is opposite to the direction of speed)

Now, Force = mass × acceleration

Force = - 1000 × 4

Force = - 4000 N (negative means that it is opposite to the direction of speed)

Hence, the unbalanced force is 4000 Newtons.

Answer 2

given m=1000kg

u=20m/s

s=20m

v=0 m/s

a=?

v square -u square =2as

20*20=2*a*50

a=4m/s