Question
Physics Question on Acceleration
The distance travelled by a particle starting from rest and moving with an acceleration 34ms−2, in the third second is
A
310m
B
319m
C
6m
D
4m
Answer
310m
Explanation
Solution
Distance travelled in the 3rd second
= Distance travelled in 3s
- distance travelled in 2 s.
As, u = 0,
S(3rds)=21a.32−21a.22=21.a.5
Given a=34ms−2;∴S(3rds)=21×34×5=310m