A small spacecraft takes off from the surface of a planet and reaches a maximum

Question

A small spacecraft takes off from the surface of a planet and reaches a maximum height then crashes. It's position at time t is given by y(t)=9t^2-4t^3 where y is measured in kilometers (km) above the surface and t is measured in minutes (min). Answer the following being careful to give correct units when called for.

a) find the speed of the spacecraft at t=1.

b) How long will it take for the spacecraft to reach its maximum height?

c) at time t=2 what is the acceleration of the spacecraft? Recalling that speed is the absolute value of velocity, is the speed increasing or decreasing at this moment?

Explanation / Answer

Position

y(t)=9t^2-4t^3

a.) t = 1

y(t) = 9 - 4 = 5 km

speed = distance / time = 5 *10^3 m / (1*60 sec) = 83.333 m/s

b.) y(t)=9t^2-4t^3

differentiate it w.r to t

y'(t) = 18t - 13t^2 = 0

t = 1.3846 sec

c.) t = 2 sec

y(t) = 9* ( 0.7692)^2 - 4(0.7692)^3

y(t) = 3.504 m/s

V = y(t)/t = 4.556 m/s

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