Question

A mass of 50 kg was moving with a velocity 400 m/s. A force of 40000 N is applied on the mass and its velocity is reduced to 50 m/s after some time. What is the distance travelled by the mass during this period?

Answer 1

Mass = 50 kg

u = 400 m/s

v= 50 m/s

F = 40000 N

t =?

Distance travelled in time t =?

a = F/m = 40000/50 = 800 m/s²

Acceleration ( a ) = change of velocity /t

⇒ t = change of velocity/a = 350/800 sec

★ Distance travelled: s = ut – (1/2)a t²

= 400 × (350/800) – (1/2) × 800 × (350/800)²

= 175 – 76.5 = 98.5 m

Answer 2

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Mass = 50 kg

u = 400 m/s

V = 50 m/s

F = 40000 N

t = unknown

Distance travelled in time (t) = ?

Acceleration (a) = F/m

= 40000/50

= 800 m/s²

Again,

Acceleration = change of velocity/t

or, t = change of velocity/a

or, t = 350/800 sec

Distance travelled:

s = ut – (½)at²

= 400 × (350/800) – (½) × 800 × (350/800)²

= 175 – 76.5

= 98.5 m