Question

Easiest question of the day







An object of mass 400gm is travelling at 5m/s. A
force of 2N is applied on it for 3 seconds in direction
that is opposite the direction of motion. So the
distance travelled by object and displacement of
object in these three seconds will be :
25/2 m and 15/2 m

But how?​

Answer 1

Given info : An object of mass 400gm is travelling at 5m/s. A force of 2N is applied on it for 3 seconds in direction that is opposite the direction of motion.

To find : The distance travelled by object and displacement of object in these three seconds will be ...

solution : m = 400g = 0.4 kg ,

initial velocity of object, u = 5 m/s

force applied on it, F = 2N

so acceleration of object, a = F/m = 2/0.4 = 5 m/s²

time , t = 3 sec

using formula, v = u + at

⇒v = 5 + -5 × 3 = -10 m/s [ a/c to question, force acts just opposite of its motion, so acceleration must be negative ]

let see, when velocity of object becomes zero,

0 = 5 - 5t ⇒t = 1sec

after 1 sec, velocity of object becomes zero after that object moves backwards.

so, distance travelled during 1 sec by object, s₁ = ut + 1/2 at²

= 5 × 1 + 1/2 × -5 × 1² = 5 - 5/2 = 5/2 m

after 1 sec, initial velocity of object , u = 0

and acceleration , a = 5 m/s² [ backwards]

so, distance covered between 1s and 3s, s₂ = 0 ÷ 2 + 1/2 × 5 × (2)² = 10 m

total distance covered by object = |s₁| + |s₂|

= 5/2 + 10 = 25/2 m

total displacement covered by object = s₂ - s₁

= 10 - 5/2 = 15/2 m [ backwards ]

I hope you got your answer, didn't you ? :)