Question

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

Answer 1

Explanation:

Displacement: d(t) = 2t^3 + 5t^2 - 3

Velocity: (d/dt)(2t^3 + 5t^2 - 3) = v(t) = 6t^2 + 10t

Acceleration: (d/dt)(d/dt)(6t^2 + 10t) = a(t) = 12t + 10

At time t = 4hr,

d(4) = [128 + 80 - 3] km = 205 km

v(4) = [96 + 40] km/hr = 136 km/hr

a(4) = [48 + 10] km per hour squared = 58 km over hours squared

Answer 2

Explanation:

hello,

for t = 4 hours

`d(4)=24^3+54^2-3=2*64+5*80-3=128+80-3=205`

this is 205 km

d is differentiable and

`d'(t)=6t^2+10t`

and

`d'(4)=6*4^2+10*4=6*16+40=96+40=136`

this is 136 km/h

d' is differentiable and

`d''(t)=12t+10`

and

`d''(4)=12*4+10=48+10=58`

this is
`58 \ km/h^2`