Question

the distance transversed by a particle moving along a straight line is given by x = 180t +50 t 2 metre . find (1) the initial velocity of a particle. (2) the velocity at the end of 4s and (3) the acceleration of the particle


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Answer 1

Answer -

Initial velocity = 180 m/s

Velocity at 4 s = 580 m/s

Acceleration = 100 m/s^2

◆ Explanation -

Displacement of particle is given by -

x = 180t + 50t^2

Differentiate w.r.t. t,

dx/dt = 180 + 50×2t

v = 180 + 100t

Acceleration of the particle is given by -

dv/dt = 0 + 100

a = 100 m/s^2

Initial velocity of particle is -

v1 = 180 + 100×0

v1 = 180 m/s

Velocity of particle at 4 s is -

v2 = 180 + 100×4

v2 = 580 m/s