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 29.0 m ahead of the red car.
Required:
a. How long after you first see the cars does the red car catch up to the blue car?
b. How far did the red car travel between when you fist saw it and when it caught up to the blue car?
c. Suppose the red car started to accelerate at a rate of a equal to 5/3 m/s^2 just at the moment you saw the cars.
d. How long after that would the red car catch up to the blue car?

Answer 1

Answer:

a) 3.66 s

b) 124.4 m

c) 3.12s

Explanation:

Given that

Speed of the Red Car, v₁ = 34 m/s

Speed of the Blue Car, v₂ = 28 m/s

Distance between the two cars, d = 22 m

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

From this, we can deduce that the red car will be catching up to the blue car at a speed of 6 m/s.

1)

If we divide the distance by the difference in speed. This becomes

d/v = 22/6 = 3.66 s. Which means, It takes 3.66 seconds for the red car to meet up with the blue car.

2

From the previous part, we were able to confirm that it took 3.66 seconds for the red car to meet up the blue car. Also, the speed with which it were traveling was travelling at, was constant, so we only need to multiply it by the time from (1) to get the distance.

v = d * t = 34 * 3.66 = 124.4

Therefore the red car travels at 124.4 m before catching up to the blue car.

3

If the red car starts to accelerate the moment we see it, the time it will take, to get to the blue car will be less than what we had gotten. We can find this using one of the equations of motion.

S = ut + ½gt², where

S = 22

u = 6

t = ?

g = 2/3

22 = 6t + 1/3t²

By using the quadratic formula, we find out the two answers listed below

t1 = 3.12 s

t2 = - 21.12 s

We all know that negative time is not possible, so the answer is t1. At 3.12 seconds