The displacement (in meters) of a particle moving in a straight line is given by s = 5/t2, where t is measured in seconds. Find the velocity of the particle at times t = a, t = 1, t = 2, and t = 3.

Answer 1

This question isn't as scary as it looks, but it does require some background in physics in addition to basic knowledge of calculus.

When you hear the word "velocity" you probably think of speed. Your speed is the rate at which your location (or position) is changing. Your displacement measures how much your position has moved. In other words, your displacement tells you your new position in terms of your distance from your original position.

With this in mind, thinking of velocity as the rate at which your position, or displacement, is changing, we get the important relationship:

velocity is the derivative of position with respect to time.

Now we can solve the problem.

Let s(t) be the displacement of the object.

Let v(t) be the velocity of the object.

It is given that s(t) = 5/(t²)

We rewrite this using properties of exponents, that an exponent in the denominator can be expressed as a negative exponent.

s(t) = 5t^-2

Next, we use the equation relating velocity to displacement.

v(t) = d/dt s(t)

v(t) = d/dt (5t^-2)

finally, we differentiate using the power rule. We multiply by the power and then lower the power by 1.

v(t) = -10 t^-3