Question:

A particle starts from the origin with uniform acceleration.Its dislacement after t seconds is given by the relation $${x=2+5t+7t^2}$$

Calculate the magnitude of its

• (1) initial velocity
• (2) velocity at t=4 sec.
• (3) uniform acceleration
• (4) Displacement at t=5 sec.

Solution:

(1) Initial velocity:

The displacement of the particle after t seconds is given by the relation

$${x=2+5t+7t^2}$$

The velocity of the particle can be found by differentiating the dislacement with respect to time.

$${v=\frac{dx}{dt}}$$

$${v=5+14t}$$

Initial velocity of the particle(at t=0) is

$${v=5 \ \frac{m}{s}}$$

The initial velocity of the particle is 5 ms^-1

(2) Velocity at t=4 sec

Velocity at t= 4 sec is

$${v=5+14t}$$

$${v=5+14(4)}$$

$${=61 \ \frac{m}{s}}$$

The velocity of the particle at t=4 sec is 61 ms^-1

(3) Uniform acceleration:

Acceleration is the rate of change of velocity

$${a=\frac{dv}{dt}}$$

$${a=\frac{d}{dt}(5+14t)}$$

$${a=14 \ \frac{m}{s^2}}$$

Uniform acceleration of the particle is 14 ms^-2

(4) Displacement of the particle at t=5 sec

$${x=2+5t+7t^2}$$

$${x=2+5(5)+7(5^2)}$$

$${202 \ m}$$

The displacement of the particle at t=5 sec is 202 m.