Distance x(t) is given by:

$\overline{){\mathbf{x}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{=}}{{\mathbf{\int}}}_{{\mathbf{t}}_{\mathbf{1}}}^{{\mathbf{t}}_{\mathbf{2}}}{\mathbf{v}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{d}}{\mathbf{t}}}$

This implies that the distance is the area under the curve.

To determine the average velocity, we proceed as follows:

x = area under the curve.

For a trapezium, A = (1/2)(a + b)h = (1/2)(8 + 2)v

21 = 5v

v = 21/5 = 4.2 m/s

**A)**

In the first 4.0s, the distance covered is:

A car in stop-and-go traffic starts at rest, moves forward 21 m in 8.0 s, then comes to rest again. The velocity-versus-time plot for this car is given in the figure.

A) What distance does the car cover in the first 4.0 seconds of its motion?

B) What distance does the car cover in the last 2.0 seconds of its motion?

C) What is the constant speed *V* that characterizes the middle portion of its motion?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Calculating Displacement from Velocity-Time Graphs concept. You can view video lessons to learn Calculating Displacement from Velocity-Time Graphs. Or if you need more Calculating Displacement from Velocity-Time Graphs practice, you can also practice Calculating Displacement from Velocity-Time Graphs practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Kosztin's class at MIZZOU.