Question

The position of a particle along x-axis at time t is given by x=1+t-t^2, the distance travelled in first 2 second is ?

Answer 1

Trajectory of motion of particle is x = 1 + t - t² , this is parabolic equation.
for finding distance first of all we have to find out crictical point where particle change the nature of motion.
So, differentiate x with respect to time
dx/dt = 0 + 1 -2t
0 = 1 - 2t ⇒t = 1/2 = 0.5
at t = 0.5 , we have to check d²x/dt² .
now, again differentiate dx/dt with respect to t
d²x/dt² = -2 < 0
Hence, at t = 0.5 , particle reaches maximum velocity and also chanhinf the nature of motion [ after t = 0.5 , particle is retardating ]

So, total distance can be found by
S = |x(t = 0.5 ) - x(t = 0)| + |x(t = 2 ) - x(t = 0.5)|
= |1 + 0.5 - 0.25 - 1 | + |1 + 2 - 4 - 1 - 0.5 + 0.25 |
= 0.75 + |-2-0.25|
= 0.75 + 2.25
= 3 m.

Hence, answer is (D)