Question

Please Help me!!!

The displacement of a particle d (in km) as a function of time t (in hours) is given by: () = 2^3 + 5^2 − 3 Find the displacement, velocity and acceleration at t = 4 hours. Indicate the correct units for each of these quantities.

Answer 1

Explanation:

The displacement of a particle d (in km) as a function of time t (in hours) is given by :

`d=2t^3+5t^2-3`

Displacement at t = 4 hours,

`d(4)=2(4)^3+5(4)^2-3=205\ km`

Velocity of particle is given by :

`v=(dd)/(dt)\\\\v=(d(2t^3+5t^2-3))/(dt)\\\\v=6t^2+10t`

Velocity at t = 4 hours,

`v=6(4)^2+10(4)=136\ km/h`

Acceleration of the particle is given by :

`a=(dv)/(dt)\\\\a=(d(6t^2+10t))/(dt)\\\\a=12t+10`

At t = 4 hours,

`a=12(4)+10=58\ km/h^2`

Therefore, the displacement, velocity and acceleration at t = 4 hours is 205 km, 136 km/h and 58 km/h² respectively.