Applied Mechanics

Q1: The equation of motion of a particle starting from rest along a straight line is x = t3 - 3t2 + 5. The ratio of the accelerations after 5 sec and 3 sec will be

A 2

B 3

C 4

D 5

ANS:A - 2

Equation of motion x = t^3-3t^2+5.

The equation to calculate the distance (in question it is given that the particle is moving along a straight line).

Differentiate with time to get velocity:

v = dx/dt = 3t^2-6t --> Velocity.
a = d^2x/dt = 6t-6 --> Acceleration.

Hence, when t = 5 sec.
a = 6*5 - 6 = 24.

When t = 3sec.

a = 6*3 - 6 = 12.
Ratio = 24/2 = 2. x = t3 - 3t2 + 5.
dx/dt = t2/2 - 3t.
putting t=5
= 12.5 - 15.
= -2.5.
putting t=3
4.5 - 9.
= -4.5.

4.5/2.5 = 1.8
~2. Equation of motion x = t^3-3t^2+5.

Equation to calculate distance (in question it is given that the particle is moving along a straight line).

Differentiate with time to get velocity:

v = dx/dt = 3t^2-6t.....Velocity.

a = d^2x/dt = 6t-6.....Acceleration.

Hence, when t = 5 sec.

a = 6*5 - 6 = 24.

When t = 3sec.

a = 6*3 - 6 = 12.

Ratio = 24/2 = 2.



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