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A ball accelerates at 8 cm/s^2 for 3 seconds, starting with velocity 12 cm/s.

What will be its velocity after the 3 seconds has elapsed?

Accelerating at 8 cm/s^2 for 3 seconds the velocity changes by

• `dv = a * `dt = 8 cm/s^2 * (3 s) = 24 cm/s

Starting at 12 cm/s and changing velocity by 24 cm/s it ends up moving at 36 cm/s, so

• vf = v0 + `dv = 12 cm/s + 24 cm/s = 36 cm/s.

Assuming that acceleration is constant, what will be its average velocity during this interval?

If acceleration is constant then the v vs. t graph is linear.

If a graph is linear its average value on any interval is the average of its initial and final values, and occurs at the midpoint of the interval.

Thus the average velocity is the average of the initial and final values:

• vAve = (12 cm/s + 36 cm/s) / 2 = 24 cm/s

How far will it travel during this interval?

The displacement during the interval is 24 cm/s * 3 s = 72 cm, which is the product of average velocity and time interval:

• `dx = vAve * `dt = 24 cm/s * 3 s = 72 cm. 

The object travels only in one direction, so the answer to 'how far' is the magnitude of its displacement, which is | 72 cm | = 72 cm. 

This last comment might seem unnecessary but note that another object might well travel well beyond its finishing point 72 cm away before turning around and coming back to it, thereby traveling much further than its displacement. 

It would also be possible for an object to remain between its initial and final points, but reverse direction several times before coming to the final point, therefore again traveling a distance greater than its displacement.

Note solution for University Physics students using integration, below the Miscellaneous Comments

Miscellaneous Comments:
 

Useful questions to ask when finding the average velocity:

The initial velocity on this interval is 12 cm/s, and the final is 36 cm/s.
The average velocity will lie between the initial and final velocities.
The average velocity is not equal to 12 cm/s or to any lesser velocity, and it is not equal to 36 cm/s or to any greater velocity.
What therefore do you think is the average velocity on this interval?
 

Incorrect student solution:  final velocity is 12 cm / s + 24 cm / s = 36 cm / s, so displacement = 36 cm/s * 3 s = 108 cm

36 cm/s is the maximum velocity on this interval. Your result indicates how far it would travel in 3 seconds, if it traveled at its maximum velocity.
However reaches its maximum velocity only at the end of the interval, so your result is an overestimate.
What quantity would you multiply by the time interval to get the displacement, or change in position?  (This question can be answered by using the definition of velocity.)

Nearly correct student solution:

 `ds = (vf +v0)/2 * `dt</p>
= (36 cm/s + 12 cm/s)/2 * 3s</p>
= 48 cm/s / 2 * 3s</p>
= 24 cm/s * 3s</p>
= 8 cm</p>

INSTRUCTOR COMMENT:

You had an arithmetic error at the very last step.  Otherwise you've done this very well.

24 * 3 = 72, not 8.

The error could have been detected by thinking about whether the result makes sense.

Observe that:

• The ball moves an average of 24 cm every second, so in 3 seconds it moves 72 cm.
• Its velocity changes by 24 cm/s in 3 sec, so every second its velocity changes by 8 cm/s, giving it average acceleration of 8 cm/s^2.

 

For University Physics Students:  Solution by Integration

Short Solution: 

• Integrate v(t) = 12 cm/s + 8 cm/s^2 * t from t = 0 sec to t = 3 sec. 

• An antiderivative is 12 cm/s * t + 4 cm/s^2 * t^2.

• The change in this antiderivative between the two clock times is the change in position. 

• The t = 0 and t = 3 sec values of this antidervative are 0 cm and 72 cm, so

• the change in the antiderivative and therefore the change in position is 72 cm - 0 cm = 72 cm.

The following is a longer version of this solution with more explanation:

Acceleration is constant, so calculus is not necessary to solve the motion problem.  You should completely understand the solution given above. 

You should also understand the calculus-based solution given here.

• The acceleration function is not always linear, and had this been the case, then the problem probably could not have been solved without the use of calculus.  The problem could in that case have been solved using steps analogous to those used below.

The solution:

If acceleration is uniform, then the velocity function is v(t) = v(0) + a t, where a is the (constant) acceleration and v(0) the velocity at clock time t = 0. 

In the present problem, if we assume that the 12 cm/s initial velocity occurs at t = 0, we have v(0) = 12 cm/s and a = 8 cm/s^2.  Our velocity function is therefore

• v(t) = 12 cm/s + (8 cm/s^2) * t.

The velocity after 3 second will then be

• v(3 s) = 12 cm/s + (8 cm/s^2) * (3 s) = 12 cm/s + 24 cm/s = 36 cm/s.

The displacement during this 3-second interval is the definite integral of the velocity function with respect to clock time, between t = 0 and t = 3 sec.

• An antiderivative of v(t) = v(0) + a t is v(0) * t + a * t^2 / 2. 

The definite integral is the change in the antiderivative function, which is

• (v(0) * (3 s) + a * (3 s)^2 / 2) - (v(0) * (0 s) + a * (0 s)^2 / 2)

Substituting for v(0) and a we have

(12 cm/s * (3 s) + 8 cm/s^2 * (3 s)^2 / 2) - (12 cm/s * (0 s) + 8 cm/s^2 * (0 s)^2 / 2) =

(36 cm + 36 cm) - (0 cm + 0 cm) = 72 cm - 0 cm = 72 cm.

This agrees with the result obtained using commonsense reasoning on this situation.