since v= d/t
would the time it takes to reach the van be
155m/30.0m/s= 5.17s?
~I don't think this is okay though since the van is moving so this just confuses me further..
Speedy Sue during is driving at 30.0m/s
and enters a 1 lane tunnel. She then observes a slow moving van 155m ahead traveling at 5.0m/s. She applies the brakes but can only accelerate at -2.0m/s^2 b/c road is wet.
a) will there be a collision? How do you know?
b) If there is a collision state how far in the tunnel and at what time the collision occurs. If not then determine the distance of closest approach btwn the car and van.
~well I know that
sue:
vi= 30.0m/s
a= -2.0m/s^2
xi= 155m
Van:
vi= 5.0m/s
I'm not sure what equation or what to do next. Is the acceleration constant? I assume so since this chap is about kinematics but it only says -2.0m/s^2 for acceleration so I guess it is..but do I need to find the xf for sue?
I think I would need to find xf but I don't have the final vf for Sue's car. Do I need to find that?
I need help on this..
Thank you
2 answers
The acceleration is assumed to be constant until the car has a velocity of zero w.r.t. the ground or when it collides with the van.
You could write down coordinates of the car and the van as a function of time and see if at some time the become equal.
A simpler way to solve this problem is as follows. Suppose you are in the van and look at the car approaching you, what do you see?
What you see is a car initially (at t = 0) at a distance of 155 m that is approaching you at t= 0 at a speed of 25 m/s, and acceleating at -2 m/s^2. The car will either collide with the van or it will not reach the van. The braking will then stop when the car is moving away from the van at a speed of 5 m/s.
This is similar to someone on a tower watching someone throwing a ball upward in his direction. Will the ball reach him? The ball has some initial velocity in the upward direction, but is accelerating in the downward direction. Of course the ball will continue to accelerate downward and not stop to accelerate toward the ground once a certain speed is reached, but that is irrelevant to the question of whether the ball will reach a certain height.
In case of throwing a ball, you know that you can solve that using conservation of energy. The initial kinetic energy of the ball is
1/2 m v^2 where v is the initial speed. At it's highest point the ball will be at rest, therefore the height it will reach is:
m g h = 1/2 m v^2 --->
h = 1/2 v^2/g
In case of the car-van problem v = 25 m/s and g is 2 m/s^2. Inserting these numbers gives:
h = 156.25 m
This means that the car will hit the van.
You could write down coordinates of the car and the van as a function of time and see if at some time the become equal.
A simpler way to solve this problem is as follows. Suppose you are in the van and look at the car approaching you, what do you see?
What you see is a car initially (at t = 0) at a distance of 155 m that is approaching you at t= 0 at a speed of 25 m/s, and acceleating at -2 m/s^2. The car will either collide with the van or it will not reach the van. The braking will then stop when the car is moving away from the van at a speed of 5 m/s.
This is similar to someone on a tower watching someone throwing a ball upward in his direction. Will the ball reach him? The ball has some initial velocity in the upward direction, but is accelerating in the downward direction. Of course the ball will continue to accelerate downward and not stop to accelerate toward the ground once a certain speed is reached, but that is irrelevant to the question of whether the ball will reach a certain height.
In case of throwing a ball, you know that you can solve that using conservation of energy. The initial kinetic energy of the ball is
1/2 m v^2 where v is the initial speed. At it's highest point the ball will be at rest, therefore the height it will reach is:
m g h = 1/2 m v^2 --->
h = 1/2 v^2/g
In case of the car-van problem v = 25 m/s and g is 2 m/s^2. Inserting these numbers gives:
h = 156.25 m
This means that the car will hit the van.
0 answers