Question

The displacement of a particularly as a function of time is given as x(t)=(6t^2-t^3)m.find the displacement,velocity and acceleration at t=0s,t=2s,t=4s,t=6s

Answer 1

Answer:

for displacement the equation is given

put the values of t in that equation

for velocity differentiate the equation of displacement

put the values of t in that equation

for acceleration differentiate the equation of velocity

put the values of t in that equation

you will get the answer

please mark me as a brainleast

Answer 2

Answer :

Displacement = x = 6t² - t³.

Differentiating x :

⇒ v = dx/dt

⇒ 2*6t - 3t²

⇒ 12t - 3t²

Now, Instantaneous velocity at 0 second :

⇒ 12*0 - 3*(0)²

⇒ 0 - 3*0

⇒ 0 m/s

Hence, Instantaneous velocity at time of 0 second = 0 m/s.

Now, Instantaneous velocity at 2 seconds :

⇒ 12*2 - 3*(2)²

⇒ 24 - 3*4

⇒ 24 - 12

⇒ 12 m/s

Hence, Instantaneous velocity at time of 2 seconds = 12 m/s.

Now, Instantaneous velocity at 4 second :

⇒ 12*4 - 3*(4)²

⇒ 48 - 3*16

⇒ 48 - 48

⇒ 0 m/s

Hence, Instantaneous velocity at time of 6 seconds = 0 m/s.

Now, Instantaneous velocity at 6 seconds :

⇒ 12*6 - 3*(6)²

⇒ 72 - 3*36

⇒ 72 - 108

⇒ -36 m/s

Hence, Instantaneous velocity at time of 6 seconds = -36 m/s.

Now, Finding Instantaneous acceleration :

⇒ 12 - 2*3t

⇒ 12 - 6t

Acceleration at 0 second ⇒ 12 - 6*0 = 12 m/s²

Acceleration at 2 seconds ⇒ 12 - 6*2 = 12 - 12 = 0 m/s²

Acceleration at 4 seconds ⇒ 12 - 6*4 = 12 - 24 = -12 m/s²

Acceleration at 6 seconds ⇒ 12 - 6*6 = 12 - 36 = -24 m/s²

Hence, Acceleration at t = 0, 2, 4, 6 seconds is 12 m/s², 0 m/s², -12 m/s² and -24 m/s².