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Home Physics Examples Newtonian Mechanics Non uniform acceleration problem

Non uniform acceleration problem

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This is a problem I discovered on the Yahoo answers physics page: A toy rocket of mass 0.2 kg is projected vertically upwards from rest by means of force which decreases uniformly in 2 seconds from 2 kg wt to zero and thereafter ceases. Find the maximum height reached.

First of all, I was not aware of the fancy kg wt unit. A simple Google search showed that 1 kg wt = 1 N. So we can now begin with a sketch of the problem.

 

The problem is divided into 2 parts: one with non uniform acceleration (represented by h_{1}, from t = 0 s to t = 2 s) and the other with uniform deceleration (h_{2}, after t = 2 s).

The non-uniform acceleration is given by

a = \frac{F}{m} - g

After 2 s, the acceleration becomes

a = -g

The (1D) force F at time t can be determined using the conditions prescribed  by the problem and recalling that F has a linear profile:

\frac{F - 0}{t - 2} = \frac{2g - 0}{0 - 2}

F = -g\left(t - 2\right)

and hence

\int_0^{v_{2s}}\,dv =\int_{t = 0 s}^{t = 2 s}\!a\,dt =\int_{t = 0 s}^{t = 2 s}\!\frac{2 - t}{0.2}g\,dt

The velocity of the rocket at t = 2 s is therefore

v_{2s} = 8g

Also,

v = \frac{dh}{dt}

We can therefore calculate h_{1}

\int_0^{h_{1}}\,dh =\int_{t = 0 s}^{t = 2 s}\!v\,dt =\int_{t = 0 s}^{t = 2 s}\!\left(9t - \frac{5}{2}t^{2}\right)g\,dt

This gives h_{1} = 34\frac{g}{3} m.

Determining h_{2} is a simple problem with uniform acceleration

0 = \left(8g\right)^{2} - 2gh_{2}

This gives h_{2} = 32g m.

Therefore, the maximum height = h_{1} + h_{2} = \frac{130g}{3} m.

Last Updated on Saturday, 14 November 2009 19:38  

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