Question

A ball starts moving eastwards with speed 35m/. An acceleration of 10 m/s^2 is acting on it in westward direction. What is the distance traveled by particle in 4^th second of its motion?
Answer is 10/4 meters

Answer 1

A ball starts moving eastward with speed 35m/s

so, velocity of ball , $\vec{u}=35\hat{i}$

an acceleration of 10 m/s² is acting on it in westward direction.

so, acceleration, $\vec{a}=-10\hat{i}$

here it is clear that, acceleration is just opposite direction of velocity of motion.

using formula, v = u + at

to find time where v = 0

0 = 35 - 10t => t = 3.5 sec

after 3.5 sec body moves in direction of acceleration.

velocity at t = 3sec

v = 35 - 3 × 10 = 5 m/s

it is initial velocity when we find distance covered in 3 sec to 3.5 sec

and initial velocity= 0, when we find distance covered in 3.5 sec to 4sec

so, distance covered in 4th second = distance covered in 3 sec to 3.5 sec + distance covered in 3.5 sec to 4 sec

= |5 × 0.5 - 1/2 × 10 × 0.5² | + |0 × 0.5 + 1/2 × 10 × 0.5² |

= | 5/2 - 5/4 | + |5/4 |

= 5/4 + 5/4 = 10/4 m

hence, distance covered in 4th second = 10/4 m