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.

Step-by-step explanation:

hello,

for t = 4 hours

this is 205 km

d is differentiable and

and

this is 136 km/h

d' is differentiable and

and

this is

Step-by-step 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

the answer is 0.3175 hope that you ^w^

also you might actually want to put this picture up if you want people to solve it luckily i remember this question

£56

step-by-step explanation:

u divide 24 by 3 and u get 8 so if u do 8 time 7 or 8+8+8+8+8+8+8 u get 56 and 56 divided by 7 is 8.