Class 11 NEET Physics Motion in a Straight Line Questions

x=40+12t−t3
∴ Velocity, v=

dx

dt

=12−3t2
When particle come to rest,

dx

dt

=v=0
∴12−3t2=0⇒3t2=12 ⇒t=2‌sec
Distance travelled by the particle before coming to rest

s

∫

0

ds=

2

∫

0

vdt or s=

2

∫

0

(12−3t2)dt =12t−

3t3

3

|02
s=12×2−8=24−8 =16m

(OR)


At t=0, particle is at , let's say x distance, from O;
then putting t=0 in the given displacement-time equation we get;
x=40+12(0)−(0)3=40m
Particle comes to rest that means velocity of particle becomes zero after travelling certain displacement ; let's say the time be t. then after differentiating the given displacement-time equation w.r.t. time we get velocity - time equation
v=12−3t2
when the particle comes to rest ):
at time t=t
v=0;
=>12−3t2=0
=>t=2s
Then, at t=2 s we are at , let's say x′ distance from O;
put this value oft (=2) in given displacement-time equation,
we get; x′=40+12(2)−(2)3
=56m
Further; We have seen that the particle started his journey when it is at 40m from the point O. And came to rest at 56m from the point O.
then the particle traveled a distance of:
56−40=16m