1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A bus starting from rest moves with a uniform acceleration of 0.1 ms−2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

A
12 ms1 , 700 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12 ms1 , 720 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
10 ms1 , 700 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10 ms1 , 720 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 12 ms−1 , 720 mGiven, initial speed of the bus, u=0, acceleration, a=0.1 ms−2 and time taken, t=2 minutes=2×60=120 s Let v be the final velocity and s be the distance covered. (a) From the first equation of motion,v=u+at, v=0+(0.1×120) ⇒ v=12 ms−1 (b) According to the third equation of motion, v2−u2=2as (12)2−(0)2=2×0.1×s ⇒s=720 m Speed acquired by the bus is 12 ms−1 and the distance travelled is 720 m.

Suggest Corrections
625
Join BYJU'S Learning Program
Related Videos
Third Equation of Motion
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program