Example from Physics, 2.4 Velocity vs. Time Graphs

Using Velocity Graph to Calculate Some Stuff: Jet Car

Use this figure to (a) find the displacement of the jet car over the time shown (b) calculate the rate of change (acceleration) of the velocity. (c) give the instantaneous velocity at 5 s, and (d) calculate the average velocity over the interval shown.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Strategy
1. The displacement is given by finding the area under the line in the velocity vs. time graph.
2. The acceleration is given by finding the slope of the velocity graph.
3. The instantaneous velocity can just be read off of the graph.
4. To find the average velocity, recall that $v_{avg} = \frac{Δ d}{Δ t} = \frac{d_{f} - d_{0}}{t_{f} - t_{0}}$
2
Solution
1. 1. Analyze the shape of the area to be calculated. In this case, the area is made up of a rectangle between 0 and 20 m/s stretching to 30 s. The area of a rectangle is length $\times$ width. Therefore, the area of this piece is 600 m.
  2. Above that is a triangle whose base is 30 s and height is 140 m/s. The area of a triangle is 0.5 $\times$ length $\times$ width. The area of this piece, therefore, is 2,100 m.
  3. Add them together to get a net displacement of 2,700 m.
2. 1. Take two points on the velocity line. Say, t = 5 s and t = 25 s. At t = 5 s, the value of v = 40 m/s.
    At t = 25 s, v = 140 m/s.
  2. Find the slope. $\begin{matrix} a & = & \frac{Δ v}{Δ t} \\ = & \frac{100 m/s}{20 s} \\ = & 5 {m/s}^{2} \end{matrix}$
3. The instantaneous velocity at t = 5 s, as we found in part (b) is just 40 m/s.
4. 1. Find the net displacement, which we found in part (a) was 2,700 m.
  2. Find the total time which for this case is 30 s.
  3. Divide 2,700 m/30 s = 90 m/s.

0 of 2 steps open

How did it go?

Strategy

Solution

Strategy

Solution