Question

You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a constant Vr equal to 34 m/s and the blue car is moving at constant Vb equal to 28 m/s. At the moment you fist see them, the blue car is 22.0 m ahead of the red car.
1)How long after you first see the cars does the red car catch up to the blue car? (Express your answer to two significant figures.) (in seconds)

2)How far did the red car travel between when you fist saw it and when it caught up to the blue car? (Express your answer to two significant figures.) (in meters)

3)Suppose the red car started to accelerate at a rate of a equal to 2/3 m/s2 just at the moment you saw the cars. How long after that would the red car catch up to the blue car? (Express your answer to two significant figures.) (in seconds)

Answer 1

Answer:

1)  3.66 s

2) 124.44 m

3) 3.12 s

Explanation:

Let's start by first listing down the information in the question.

Red Car : 34 m/s

Blue Car: 28 m/s

Distance between them : 22 m

The difference in speed between the cars is: 34 - 28 = 6 m/s

This means that the red car is catching up to the blue car at a speed of 6 m/s.

1) We can solve this by just dividing the distance by the difference in speed. This becomes:    $\frac{Distance}{Speed}= \frac{22}{6} = 3.66$

Thus it takes 3.66 seconds for the red car to catch up to the blue car.

2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.

This becomes:    $Speed * Time = 34 * 3.66 = 124.44$

Thus the red car travels 124.44 m before catching up to the blue car.

3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.

We can use the motion equation : $s = u*t + \frac{1}{2}(a * t^2)$

Here s = 22 m

We can take u as the difference in speed. u = 6 m/s

Acceleration a = (2/3) m/s^2

Substituting the these into the equation we get:

$22 = 6t + \frac{1}{2}(\frac{2}{3}t^2)$

Solving this for the variable 't' using the quadratic formula we get the following two answers:

t1  = 3.12 s

t2 = - 21.12 s

Since t2 is not possible, the answer is t1. This means it takes 3.12 seconds for the red car to catch up to the blue.