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Chapter 2: Problem 64

If position (in meter) of a particle moving in straight line is given by \(x=t^{2}-2 t+1\) (where \(t\) is time in second). The distance travelled by particle in first two second is (A) Zero (B) \(2 \mathrm{~m}\) (C) \(4 \mathrm{~m}\) (D) \(3 \mathrm{~m}\)

Short Answer

Expert verified
The distance travelled by the particle in the first two seconds is (A) Zero.

Step by step solution

01

Calculate the position at t=0 sec.

Substitute t=0 in the given equation: \(x_0 = (0)^2 - 2(0) + 1\) \(x_0 = 1 \mathrm{~m}\)
02

Calculate the position at t=2 sec.

Substitute t=2 in the given equation: \(x_2 = (2)^2 - 2(2) + 1\) \(x_2 = 4 - 4 + 1\) \(x_2 = 1 \mathrm{~m}\)
03

Calculate the distance travelled in the first two seconds.

We know the initial and final position of the particle, and we can find the distance between them: Distance travelled = \(|x_2 - x_0|\) Distance travelled = \(|1 - 1|= 0\mathrm{~m}\) After analyzing the given options, the correct answer is (A) Zero.

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