Question

x= 3-5t +2t^2 find it velocity at t =2s and acceleration at t =4s

Answer 1

x=2 t^2-5 t+3 (given)

Velocity =4 t-5 ;

Therefore velocity at t=2 s =8-5 =3 m/s

Acceleration =0m/s^2

Answer 2

Answer :

Displacement = x = 3 - 5t + 2t².

Differentiating x :

⇒ v = dx/dt

⇒ 0 - 5 + 2*2t

⇒ 4t - 5

Now, Instantaneous velocity at 2 seconds :

⇒ 4*2 - 5

⇒ 8 - 5

⇒ 3 m/s

Hence, Instantaneous velocity at time of 2 seconds = 3 m/s.

Now, Instantaneous acceleration : 4 - 0 = 4 m/s².

Some Additional Information -

• Distance is the total length of the path travelled by a body.

• Displacement is the shortest distance between the initial position and the final position of the body.

• Distance is equal or greater than displacement.

• Distance is a scalar quantity and displacement is a vector quantity.