Question

A particle starts with an initial velocity of 10 m/s towards north. It experience a constant acceleration of 4 m/s^2 towards south. The distance travelled by the particle in 3rd second is ?

Answer 1

Explanation:

given u =10m/s

retardation a=4m/s²

t=3sec

final velocity v=u+at

• v=10+(-4)*3
• v=10-12=-2m/s
• that will 2 m/stowards south

Answer 2

Given: Particle starts with initial velocity=10m/s towards north. It experiences constant acceleration of 4m/s^2 towards south.

To find: Distance travelled in 3rd second

Explanation: Since the object is travelling with 10m/s towards north and acceleration is towards south, the acceleration retards the body.

The acceleration will be negative in this case. We can write acceleration as a= -4m/s^2

Object velocity (u)= 10m/s

The distance travelled by body in nth second is given by formula:

$distance = u + \frac{a}{2} (2n - 1)$

Therefore, distance travelled in 3rd second is:

=> distance= 10 + -4/2 (2*3-1)

= 10 + -2*5

= 0 m

Therefore, the object travels 0 m in third second of its motion.