Question

The distance travelled by a particle moving along a straight line is given by X= 180t + 50t square metres . find the initial velocity of the particle . The velocity at the end of 4 second and third the acceleration of the particle

Answer 1

Concept:

We have to differentiate the given equation for distance to calculate the given properties.

Given:

Distance travelled by a particle in a straight line is X = 180t + 50t²m

t = 4secs

Find:

We need to determine:

i. the initial velocity of the particle,

ii. the velocity at the end of 4second and

iii. the acceleration of the particle

Solution:

It has been given to us that the distance, X = 180t + 50t²m

differentiating the above equation w.r.t time

v = dx/dt = 180 + 502t

v = 180 + 100t

Therefore, V_initial at t = 0 = 180m/s

Therefore velocity at end of 4 seconds becomes-

v = 180 + 100 × 4

v = 180 + 400

v = 580m/s is the velocity at the end of 4sec

We have the equation, v = 180 + 100t

differentiate the above equation w.r.t. time

dv/dt = 0 + 100

We know, dv/dt = a

Therefore, a = 100 m/s²

Thus, the initial velocity of the particle is 180m/s, the velocity of the particle at the end of 4 seconds is 580m/s and the acceleration of the particle is 100m/s².

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