A model rocket rises with constant acceleration to a height of 3 m,at which point its speed is 25 m/s. How much time does it take for the rocket to reach this height and what is the magnitude of the rocket's acceleration?
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An object with an initial velocity u m/s, accelerating at a m/s^2 for a time period t has a velocity v = u + a*t after t seconds. The distance traveled by the object in t seconds is given by s = u*t + (1/2)*a*t^2.
In the problem the initial velocity of the rocket when it takes off is 0. At a height of 3 m its speed is 25 m/s. Let T denote the time taken by the rocket to reach this height and A denote its acceleration.
This gives: 25 = 0 + A*T and 3 = 0*T + (1/2)*A*T^2
Substituting 25 = AT
=> 3 = (1/2)*25*T
=> T = 6/25
A = 625/6 m/s^2
The rocket takes 6/25 seconds to reach the height and its acceleration is 625/6 m/s^2.
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