College Physics 2.66 – Solving for velocity and acceleration from position graph


Figure 2.68 shows the position graph for a particle for 6 s.

(a) Draw the corresponding Velocity vs. Time graph.

(b) What is the acceleration between 0 s and 2 s?

(c) What happens to the acceleration at exactly 2 s?

position graph for a particle for 6 s.
Figure 2.68

Solution:

Part A

The velocity of the particle is the slope of the position vs time graph. Since the position graph is compose of straight lines, we can say that the velocity is constant for several time ranges.

Time RangesSlope of the position Graph/Velocity
0 to 2 seconds=\frac{2-0}{2-0}=1\:\text{m/s}
2 to 3 seconds=\frac{-3-2}{3-2}=\frac{-5}{1}=-5\:\text{m/s}
3 to 5 seconds0 \text{m/s}
5 to 5 seconds=\frac{-2-\left(-3\right)}{6-5}=\frac{1}{1}=1\:\text{m/s}

Based from the data in the table, we can draw the velocity diagram

velocity vs time graph

Part B

Since the velocity is constant between 0 seconds and 2 seconds, we say that the acceleration is 0.

Part C

Since there is a sudden change in velocity at exactly 2 seconds in a very short amount of time, we say that the acceleration is undefined in this case.


College Physics 2.65 – Velocity graph of a world-class sprinter


A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See Figure 2.67).

(a) What is his average velocity for the first 4 s?

(b) What is his instantaneous velocity at t=5 s?

(c) What is his average acceleration between 0 and 4 s?

(d) What is his time for the race?

A graph of  v(t)  is shown for a world-class track sprinter in a 100-m race.
Figure 2.67

Solution:

Part A

To find for the average velocity over the straight line graph of the velocity vs time shown, we just need to locate the midpoint of the line. In this case, the average speed for the first 4 seconds is

\text{v}_{\text{ave}}=6\:\text{m/s}

Part B

Looking at the graph, the velocity at exactly 5 seconds is 12 m/s.

Part C

If we are given the velocity-time graph, we can solve for the acceleration by solving for the slope of the line.

Consider the line from time zero to time, t=4 seconds. The slope, or acceleration, is

\text{a}=\text{slope}=\frac{12\:\text{m/s}-0\:\text{m/s}}{4\:\text{s}}=3\:\text{m/s}^2

Part D

For the first 4 seconds, the distance traveled is equal to the area under the curve.

\text{distance}=\frac{1}{2}\left(4\:\sec \right)\left(12\:\text{m/s}\right)=24\:\text{m}

So, the sprinter traveled a total of 24 meters in the first 4 seconds. He still needs to travel a distance of 76 meters to cover the total racing distance. At the constant rate of 12 m/s, he can run the remaining distance by

\text{t}=\frac{\text{distance}}{\text{velocity}}=\frac{76\:\text{m}}{12\:\text{m/s}}=6.3\:\sec

Therefore, the total time of the sprint is

\text{t}_{\text{total}}=4\:\sec +6.3\:\sec =10.3\:\sec