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
