A rocket of mass m = 500 kg is travelling in a straight line for a short time. T

Question

A rocket of mass m = 500 kg is travelling in a straight line for a short time.
The distance in metres covered by the rocket during this time is described
by the function

s(t) = 220t - 15t2 - 60 ln (t + 1)

where t > 0 is the time in seconds.

(a) Find a function that describes the speed of the rocket.
(b) What is the distance covered by the rocket by time t = 6 seconds?
(c) Find the value of time t when the speed of the rocket is 100 ms-1
(d) Find a function that describes the acceleration of the rocket.
(e) Find the acceleration of the rocket at t = 3 seconds.
(f) Find the time when the rocket's acceleration is -15 ms-2

Explanation / Answer

(a) The function that describes the speed of rocket is given by s'(t)

s'(t) = v = 220-30t-60/(t+1)

(b) At t=6, the distance is given by s(6)

s(6) = 220(6)-15(36)-60ln(7) = 1320-540-116.755 = 663.25

(c) The time is given by solving v= 100

220-30t-60/(t+1) = 100 =>120 = 30t+60/(t+1) => 4 = t+2/(t+1)=> 4t+4=t2+t+2 => t2-3t-2=0 => So t=1 and at t=2, the speed of rocket is 100ms-1

(d) Acceleration of the rocket is given by s"(t)

s"(t) = v'(t) = a(t) = -30+60/(t+1)2

(e) The acceleration of rocket at t=3 seconds is given by a(3)

a(3) = -30+60/(4)2 = -30+60/16 = -30+15/4 = -30+3.75 = -26.25

(f) The time for which rocket's acceleration is -15ms-2 is given by solving a(t) = -15

-30+60/(t+1)2 = -15 => 60/(t+1)2 = 15 => (t+1)2 = 60/15 = 4 => t+1 = 2, -2 => t=1,-3

But we discard t=-3 and so we get t=1, the acceleration = -15

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