physics question 16b

Question

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 ss of its motion, the vertical acceleration of the rocket is given by ay=(2.70m/s3)tay=(2.70m/s3)t, where the +y+y-direction is upward.

What is the speed of the rocket when it is 335 mm above the surface of the earth?

Answer 1

First, let us find out the location h and the speed v of the rocket as function of time t during the first 10 s.

a = dv / dt = ( 2.70 m / s³ ) t = j t , where j = 2.70 m / s³

v - v₀ = Integral a dt = Integral j t dt = j t² / 2, 

h - h₀ = Integral v dt = v₀ t + Integral j t² / 2 dt = v₀ t  + j t³ / 6

The rocket starts at rest from the ground level, therefore

  v₀ = 0 and  h₀ = 0

h(t) = j t³ / 6 for 0 < t < 10 s

h(10 s) = (2.70 m / s ³ )× (10 s )³ / 6 = 450 m > 335 m

Thus it will take less than 10 s for the rocket to reach 335 m above the Earth surface, hence we can use the above formulas for h(t) and v(t)

Now let us find out how long will it take the rocket to reach 335 m above the Earth surface

h = j t³ / 6 = 335 m

t(h) = ( 6 h / j )^1/3 = ( 6 × 335 / 2.70 )^1/3 s = 9.06 s < 10 s 

and 

v (t) = j t² / 2 = j ( 6 h / j ) ^2/3 / 2 = ( 4.5 h² j )^1/3 =

= (4.5 × 335² × 2.70)^1/3 m/s = 111 m/s