Velocity and Acceleration Examples



Example 1.

I stand next to a chair at first. I walk 5 m to the right. Then I walk 7 m to the left. Find

(a) my displacement in each direction,
(b) my total displacement,
(c) the total distance I walked, and
(d) my final distance from the chair.

Take the right side as the positive direction.

Answer.

(a) When I first walk to the right, my displacement is +5 m, since right side is positive.

When I then walk to the left, my displacement is -7 m, since left side is negative.

(b) My total displacement is my distance and direction from the start position, i.e. the chair. You can guess that it is 2 m to the left.

Or you can add the two displacements above: (+5) + (-7) = -2 m. The minus sign tells us that it is to the left.

(c) The total distance I walked is 5 + 7 = 12 m, since distance has no directions.

(d) This is the straight line distance from the chair, not the actual distance I walked. So it is 2 m, as we know from (b).



Example 2.

I stand next to a chair. I run 10 m to the right in 5 s. Then I run 10 m to the left in 5 s. Find
(a) my velocity when I run to the right,
(b) my velocity when I run to the left,
(c) my average velocity in 10 s, and
(d) my average speed in 10 s.

Take the left side as the positive direction.

Answer.

(a) I run 10 m to the right in 5 s. My speed = 10/5 = 2 m/s. Since right side is negative, my velocity is -2 m/s.

(b) I run 10 m to the left in 5 s. My speed = 10/5 = 2 m/s. Since left side is positive, my velocity is +2 m/s.

(c) In the 10 s, I have come back to the starting point next to the chair. My total displacement is zero, so my average velocity is zero.

(d) In the 10 s, the total distance I have run is 10 + 10 = 20 m. So my average speed is 20/10 = 2 m/s.



Example 3.

I throw a ball upwards at a speed of 10 m/s. After 1 s, it reaches a height of 5 m, where it stops for an instant. After 1 more second, it hits my hand at a speed of 10 m/s. Find

(a) the velocity when it leaves my hand,
(b) the velocity when it hits my hand,
(c) the acceleration when it goes up, and
(d) the acceleration when it comes down.

Take the upward direction as the positive direction.

Answer.

(a) Since up is positive, the ball leaves my hand at a velocity of +10 m/s.

(b) Since down is negative, the ball hits my hand at a velocity of -10 m/s.

(c) When it goes up, the initial velocity is u = +10 m/s when it leaves my hand. The final velocity is v = 0 m/s, when it reaches the top. The time taken is t = 1 s. So the acceleration is

a = (final - initial velocity) / time taken

a = (v - u) / t = (0 - 10) / 1 = -10 m s-2.

(d) When it comes down, the initial velocity is u = 0 m/s when it is at the top. The final velocity is v = -10 m/s, when it hits my hand. The time taken is t = 1 s. So the acceleration is

a = (final - initial velocity) / time taken

a = (v - u) / t = (-10 - 0) / 1 = -10 m s-2.

The acceleration is negative, which means it is downwards, since up is positive.

Notice that the acceleration is downwards whether the ball is going up or down. This is because when the ball goes up, it is slowing down. So an upward deceleration means a downward acceleration.


Copyright 2010 by Kai Hock. All rights reserved.
Last updated: 21 July 2011.