Question

At a particle is moving with a uniform acceleration its position time t seconds is given in metre by the relation x =3 +4t +5^2.calculate the magnitude of its initial velocity. ​

Answer 1

Answer:

Velocity of particle = (10t + 4) m/s

Explanation :

Given :

Displacement of particle is given by

x = 3 + 4t + 5t^2

=> x = 5t^2 + 4t + 3

Differentiate w. r. to t

dx/dt = d(5t^2)/dt + d(4t)/dt + d(3)/dt

dx/dt = 5(dt^2/dt) 4 (dt/dt) + 0

dx/dt = 5*2t(dt/dt) + 4(1)

dx/dt = 10(t) + 4 = 10t + 4

=> v = (10t + 4) m/s.

........................[ velocity (v) = dx/dt]

Answer 2

Explanation:

Given Displacement of particles :

X = 5t^² + 4t + 3

Differ. w.r.t. to time ( t )

dx / dt = d ( 5t^² + 4t + 3 ) / dt

= d ( 5t^² ) / dt. + d ( 4t ) / dt + d ( 3 ) / dt

= 10 t + 4 + 0 = 10t + 4

Velocity = dx / dt

So when time ( t ) the initial velocity of particles is..

Velocity = ( 10t + 4 ) meter / second

Thank you