How to Find Acceleration With Constant Velocity

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People commonly use the word acceleration to mean increasing speed. For example, the right pedal in a car is called the accelerator because its the pedal that can make the car go faster. However in physics, acceleration is defined more broadly specifically, as the rate of change of velocity. For example, if velocity changes linearly with time, like v(t)=5t miles per hour, then the acceleration is 5 miles per hour-squared, since that is the slope of the graph of v(t) against t. Given a function for velocity, the acceleration can be determined both graphically and using fractions.

Graphic Solution

    Suppose the velocity of an object is constant. For example, v(t)=25 miles per hour.

    Graph this velocity function, measuring v(t) with the vertical axis and time t with the horizontal axis.

    Note that since the graph is flat, or horizontal, its rate of change with respect to time t is therefore zero. Since acceleration is the rate of change of velocity, the acceleration in this case must be zero.

    Multiply by the radius of the wheel, if you also want to determine how far the wheel traveled.

Fractional Solution

    Form a ratio of the change in velocity over some period of time divided by the length of the period of time. This ratio is the rate of change of the velocity, and therefore is also the average acceleration over that period of time.

    For example, if v(t) is 25 mph, then v(t) at time 0 and at time 1 is v (0)=25mph and v(1)=25mph. The speed doesnt change. The ratio of the change in speed to the change in time (i.e. the average acceleration) is CHANGE IN V(T) / CHANGE IN T = [v(1)-v(0)]/[1-0]. Clearly this equals zero divided by 1, which equals zero.

    Note that the ratio calculated in step 1 is just the average acceleration. However, you can approximate the instantaneous acceleration by making the two points in time at which the velocity is measured as close as you want.

    Continuing with the example above, [v(0.00001)-v(0)]/[0.00001-0] = [25-25]/[0.00001] = 0. So clearly, the instantaneous acceleration at time 0 is zero miles per hour-squared as well, while the velocity remains a constant 25 mph.

    Plug in any arbitrary number for the points in time, making them as close as you like. Suppose they are only e apart, where e is some very small number. Then you can show that the instantaneous acceleration equals zero for all time t, if the velocity is constant for all time t.

    Continuing with the example above, [v(t+e)-v(t)]/[(t+e)-t] = [25-25]/ e = 0/e = 0. e can be as small as we like, and t can be any point in time we like, and well still get the same result. This proves that if the velocity is constantly 25 mph, then the instantaneous and average accelerations at any time t are all zero.

References

  • "Fundamentals of Physics"; David Halliday and Robert Resnick; 1992

About the Author

Paul Dohrman's academic background is in physics and economics. He has professional experience as an educator, mortgage consultant, and casualty actuary. His interests include development economics, technology-based charities, and angel investing.

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