Question

the graph above shows the velocity v as a function of time t for an object moving in a straight line. which of the following graphs shows the corresponding displacement x as a function of time t for the same time interval?
questions 4 - 5
at time t = 0, car x traveling with speed v0 passes car y, which is just starting to move. both cars then travel on two parallel lanes of the same straight road. the graphs of speed v versus time t for both cars are shown above.
which of the following is true at time t = 20 seconds?
(a) car y is behind car x. (b) car y is passing car x. (c) car y is in front of car x.
(d) both cars have the same acceleration. (e) car x is accelerating faster then car y.

1. from time t = 0 to time t = 40 seconds, the areas under both curves are equal. therefore, which of the following is true at time t = 40 seconds?

(a) car y is behind car x. (b) car y is passing car x. (c) car y is in front of car x.
(d) both cars have the same acceleration. (e) car x is accelerating faster than car y.

Explanation:

Step1: Recall the relationship between velocity - time graph and displacement

The displacement of an object is given by the area under the velocity - time graph.

Step2: Calculate the displacement of car X at t = 20 s

Car X has a constant velocity \(v_0\). The displacement of car X, \(x_X=v_0\times20\) (since the area under the velocity - time graph of car X is a rectangle with base 20 s and height \(v_0\)).

Step3: Calculate the displacement of car Y at t = 20 s

The velocity - time graph of car Y is a triangle from \(t = 0\) to \(t=20\) s. The area of a triangle is \(A=\frac{1}{2}bh\). Here, the base \(b = 20\) s and the height \(h = v_0\). So the displacement of car Y, \(x_Y=\frac{1}{2}\times v_0\times20 = 10v_0\).
Since \(x_X>x_Y\), car Y is behind car X at \(t = 20\) s.

Step4: Analyze the situation at t = 40 s

The area under the velocity - time graph represents displacement. Given that the areas under both curves from \(t = 0\) to \(t=40\) s are equal, the displacements of the two cars are equal at \(t = 40\) s. This means that car Y is passing car X at \(t = 40\) s.

Answer:

1. A. Car Y is behind car X.
2. B. Car Y is passing car X.