Question
This question is related to the May 16 question. Suppose a second car is moving along the same straight line such that its position x at time t is given by the equation x = 10t + 60, where t is measured in seconds and x in meters. At what time and at what position do the two cars pass each other?
Solve this problem both analytically, i.e., by means of a calculation, and graphically.
Answer
The equations for the position vs. time for the two cars are as follows. (The equation for the car in the May 16 question is derived in the answer to that question.)
At the instant when the two cars pass each other, both cars have the same x-values and the same t-values. Therefore, we can find the common x and t by equating the x in equation (1) to the x in equation (2) and similarly for t.
Thus, the left-hand sides of equations (1) and (2) are equal. Therefore, the right-hand sides must be equal as well. This gives
Moving the terms with t to the left-hand side, by adding 20t to both sides of the equation, and moving the constant terms to the right-hand side, by subtracting 60 from both sides of the equation, gives
Dividing both sides of equation (5) by 30 gives us t,
Substituting this value of t into equation (1) gives
You should check that you get the same value for x if you substitute value (6) for t into equation (2). This is a consistency check that shows that we did not make a silly mistake in our calculation.
Another check that the answers (6) and (7) are correct is obtained by graphing the two functions (1) and (2). The intersection point of the two straight lines is the point where both lines have the same x-values and the same t-values. Thus, the intersection point is the point where the two cars are passing each other. As you see from the graph below, this point does indeed have the coordinates (6) and (7).