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Revision:Motion with Variable Acceleration

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics Revision Notes > Motion with Variable Acceleration


Motion with Variable Acceleration

In Mechanics 1, you often learn the 'suvat' equations which will give the oppurtunity to work out the distance, speed travelled or the time taken (among other things). However, it is important to remember a limitation of these formulae: They only work when acceleration (a) is constant. As you move into more complicated areas, it is important to extend these.

In a Straight Line

Consider a particle moving along a straight line. Often a formula will be given for the distance away from a certain point it is. We can call this s = f(t), where t is the time elapsed.

Clearly f'(t) will be the rate of change of distance, namely speed. Similarly, f''(t) is the rate of change of speed. The way this is normally written is:

a = \dot{v} = \ddot{s}

v = \dot{s} \\

Where \dot{s} means the derivative of s with regard to time.

This also works in reverse, namely:

\int a \text{ d}t = v + C

\int v \text{ d}t = s + C\\

The exact values can't be worked out without some values to substitute in and eliminate the arbitary constant.

In two dimensions

Often, a displacement, velocity or acceleration will not be given by a scalar, but as a vector in terms of \mathbf{i} and \mathbf{j}. In this case, you follow the same procedure as before, only for each component part.

\mathbf{s} = f(t)\mathbf{i} + g(t)\mathbf{j}

\mathbf{v} = \dot{\mathbf{s}} = f'(t)\mathbf{i} + g'(t)\mathbf{j}

\mathbf{a} = \dot{\mathbf{v}} = \ddot{\mathbf{s}} = f''(t)\mathbf{i} + g''(t)\mathbf{j}\\

When dealing with integration, it is important to remember the arbitary constant, and there will be one of these for each component vector:

\mathbf{a} = f(t)\mathbf{i} + g(t)\mathbf{j}

\mathbf{v} = \int \mathbf{a} \text{ d}t = \int f(t) \text{ d}t \;\mathbf{i} + \int g(t) \text{ d}t\; \mathbf{j} + C_1 \mathbf{i} + C_2 \mathbf{j}\\

This can often be equated using one set of values since,  a\mathbf{i} + b\mathbf{j} = c\mathbf{i} + d\mathbf{j} \Leftrightarrow a = c and b = d