Position of a particle changes as x=(t^2) +3t+5.Find velocity and acceleration when the motion starts.

Open in App

Solution

$x = t^{2} + 3 t + 5 v e l o c i t y = \frac{d x}{d t} = 2 t + 3 a t t = 0, v = 3 m / s a c c e l e r a t i o n = \frac{d v}{d t} = 2 m / s^{2} a c c e l e r a t i o n i s u n i f o r m .$

Suggest Corrections

Similar questions

The time variation of the position of a particle in rectilinear motion is given by $x = 2 t^{3} + t^{2} + 2 t .$ If v is the velocity and a the acceleration of the particle in consistent units, the motion started with

Q. The position

x

of a particle varies with time

t

according to the relation

x = t^{3} + 3 t^{2} + 2 t

. Find velocity and acceleration as functions of time.

A particle starts its motion from point A in the positive x direction in such a way that after every t sec it is at a distance of $t^{2}$ from the starting point. Find its instantaneous velocity at $t = 4^{t h}$ sec also the rate at which velocity changes.

Q. Acceleration of the particle in a rectilinear motion is given as

a = 4 t + 3

(m / s^{2})

. The velocity and displacement as a function of time

t

respectively are
(Given that at

t = 0, u = 3 m / s, x = 0 m

)

Q. The position of a particle moving along x-axis given by

x = (- 2 t^{3} + 3 t^{2} + 5) m

. The acceleration of particle at the instant its velocity becomes zero is

Join BYJU'S Learning Program

Position of a particle changes as x=(t^2) +3t+5.Find velocity and acceleration when the motion starts.

x = t2+3t+5velocity = dxdt = 2t+3at t =0, v = 3m/sacceleration = dvdt = 2 m/s2acceleration is uniform.

$x = t^{2} + 3 t + 5 v e l o c i t y = \frac{d x}{d t} = 2 t + 3 a t t = 0, v = 3 m / s a c c e l e r a t i o n = \frac{d v}{d t} = 2 m / s^{2} a c c e l e r a t i o n i s u n i f o r m .$