displacement of particle s (in km) as function of time t (in hrs) is given by s(t)=2t^3-24t.What is acceleration when time is 30 minutes.Whatis average rate of change in velocity during first 4 hours?

Question

What is instantaneous rate of change in displacement when time is 60 minutes?

rate of change in displacement when time is 60 minute

Answer 1

s(t) = 2t³ - 24t

s'(t) = 6t² - 24

s"(t) = 12t

What is acceleration when time is 30 minutes.

Acceleration at time t (in hours) is s"(t).

s"(1/2) = 6 km/h²

Whatis average rate of change in velocity during first 4 hours?

Velocity at time t is s'(t).

During the first 4 hours velocity change was s'(4) - s'(0).

Therefore the average rate of change is

(s'(4) - s'(0)) / 4 = ((6 × 4² - 24) - (-24)) /4 = 24 km/h²

What is instantaneous rate of change in displacement when time is 60 minutes?

Rate of change in displacement at time t (in hours) is s'(t). At time 60 minutes:

s'(1) = 6 × 1² - 24 = -18 km/h

That is at time 60 minutes the particle is moving backwards at the speed of 18 km/h