Question

Please Help me!!!

Answer 1

Step-by-step 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 :

$v=\dfrac{dd}{dt}\\\\v=\dfrac{d(2t^3+5t^2-3)}{dt}\\\\v=6t^2+10t$

Velocity at t = 4 hours,

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

Acceleration of the particle is given by :

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6t^2+10t)}{dt}\\\\a=12t+10$

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.