If for a given pair of media CR
1 answer:
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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:
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:
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 :
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:
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.
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where is the initial velocity.
(a).
When the projectile hits the 50m mark, ; therefore,
solving for we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed of