Question

The distance traversed by a particle moving along a straight line is given by X=180t+50t square. Find:the initial velocity of the particle,the velocity at the end of 4seconds and the acceleration of the particle

Answer 1

Dear Student,

◆ 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

Hope this helps you...

Answer 2

Answer:

ExplanationDear Student,

◆ 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=580m/s