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

# A car accelerates from rest at a constant rate of 2 ms−2 for some time. Then it retards at a constant rate of 4 ms−2 and come to rest. It remains in motion for 6 s.

A
Its maximum speed is 8 ms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
It maximum speed is 6 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
It travelled a total distance of 24 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
It travelled a total distance of 18 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C It travelled a total distance of 24 ma1=2 ms2,a2=−4ms2 v0=a1t1=2t1,v0=a2t2=4t2 t1+t2=6⇒v02+v04=6 ⇒v0=8 ms Distance during constant accelerated motion can be calculated as Vavgt. So, AC=Vavgt1=v0+02t1 CB=Vavgt1=v0+02t2 Total distance travelled S=AC+CB=12v0t1+12×v0t2 ⇒S=12v0[t1+t2]=12×8×6=24 m

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Motion Under Constant Acceleration
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program