A particle has initial vclocity 10 m/s. It moves due to a constant force along
the line of velocity which initially produces retardation of 5 m/s^2. Then the:
(A) distance travelled in first 3 second is 10.0 m
(B) distance travelled in first 3 second is 7.5 mm
(Ci distance travelled in first second is 12.5 m
(D) distance travelled in first 3 second is 17.5 m
Answers
Answered by
1
Answer:
(A) distance travelled in first 3 second is 10.0 m
Answered by
1
Answer:
answer is C option
Explanation:
becauseinitial velocity equals to 10 metre per second and as acceleration equals to - 5 metre per second square because of retardation minus sign we have to find distance in time travel in 3 second then and s equals to to UT plus half a t square then and s equals to 30 minus half x 5 x 25 equals to to 12.5
Similar questions