Question

16.
A particle starts from rest, accelerates at 2 m/s2
for 10 s and then goes with constant speed for
30 s and then decelerates at 4 m/s2 till it stops.
What is the distance travelled by it.​

Answer 1

Answer:

For first 10 sec

distance s1 = 1/2 at^2

1/2×2×10×10 = 100m

And velocity reached=

v =u+at

v= 0+2×10 = 20m/s

For next 30s

s2= 20× 30 = 600m

Then under retardation of 4m/s^2

v^2 = 2as3

20× 20 = 2×4×s3

S3 = 50m

Total s1+ s2+ s3

100+600+50 = 750m