Solving problems on acceleration



The biggest problem about acceleration problems, is getting the direction right. It is useful to sort this out right at the start.

We shall mainly look at motion along a straight line. A body moving this straight line has two possible directions. This could be left or right, up or down, or any way you choose to think about it. In an equation like

s = ut + at2/2
the direction of the motion is indicated by the sign of the numbers s, u and a.

Lets start with a ball moving from left to right. In any problem we must first choose a position to be the origin. This is the point where we say that s = 0. It can be any point, so we try to choose something convenient, like the starting point. Next, we have to decide whether we want the positive and negative values of the displacement s to mean left or right. Either way is all right, as long as we keep to it in the problem.

Likewise, the velocity u can be positive or negative. This must follow the same way that is chosen for the displacement. If right is positive for displacement, then a ball moving to the right has positive velocity. Be careful that velocity is different from displacement. Even if displacement is negative, velocity can still be positive or negative. That is to say, where ever the ball is, it can be either direction.

The issue of directions and signs become the most confusing when it comes to the acceleration a. The reason is that it very easy to assume that acceleration and velocity have the same direction. To see that this is not true, always remember this example:

If the body is slowing down, then acceleration is opposite in direction and sign to velocity.

If the body is speeding up, then acceleration is the same in direction and sign to velocity.

For completeness, I include the following example where left is positive:

Once the signs and directions are clear, we can use the equations more confidently. Lets work out the following problems.

Problem 1.
A ball is at first moving to the right with a speed of 5 m/s. It accelerates to the left with an acceleration of 1 m/s2. what is its velocity after 10 s?

Solution.
How can a ball move to the right and accelerate to the left at the same time? Well, if it accelerates in the opposite direction to the velocity, that just means that it is slowing down.

To decide which equation to use, look for the given information:
initial velocity u= 5 m/s, acceleration a = 1 m/s2, time t= 10 s.
The required answer is the final velocity v. We need an equation that contains s, a, t, and v. That would be v = u + at.

Before substituting, remember the very important step: decide which direction to choose as the positive. Lets choose right to be positive. Then we must put in the correct signs for the given information:
u= +5 m/s, a = -1 m/s2, t= 10 s.
Now we can safely substitute into the equation v = u + at. This gives

v = 5 - 1 x 10 = -5.

So the answer is -5 m/s. We must not forget at the end to state the physical meaning of the sign: the ball moves to the left with a speed of 5 m/s.



Problem 2.
A ball is thrown upwards with a speed of 5 m/s. When it falls back down to the same height, what is its speed? How long does it take? Assume that the acceleration of free fall is approximately 10 m/s2

Solution.
When it falls back to the same height, the displacement s = 0. We know the acceleration a, the initial velocity u, and we want to find the final velocity v. The equation that contains all these items is

v2 = u2 + 2as.

Before substituting, remember to choose a direction to be positive and put in the correct signs. Lets choose up to be positive. Then
s = 0 m
a = -10 m/s2
u = +5 m/s
Now we can substitute into the above equation,

v2 = 52 + 2(-10)(0).

So v = +5 m/s or -5 m/s. Which is it? We have to think outside the equation. When the ball goes up, it must slow down because of gravity. It is not possible that after some time, the ball can still be going up at 5 m/s. So only -5 m/s is physically possible. That is, the final velocity is 5 m/s downwards.

Next, we need to find the time t. We have the values for u, v, a and s. We can use any one of the other equations. If you try them all out, you would find that the easiest one is

a = (v - u)/t.
Making t the subject:
t = (v - u)/a
and substituting
t = (-5 - 5)/(-10)
gives a time of 1 s.



Copyright 2010 by Kai Hock. All rights reserved.
Last updated: 29 November 2010.