Question

47. The position of a particle along x-axis at time t is given by x=1+t-t2. The distance travelled by the particle in first 2 seconds is ?​

Answer 1

Answer:

=0+1−2t

0=1−2t

⇒t=

2

1

=0.5

At t=0. , we have to check $$\dfrac{d^2x}{dt^2}.

Now, again differentiate $$\dfrac{dx}{dt} with respect to t

dt

2

d

2

x

=−2 Which is <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

=3m.