1) One drop of oil falls straight down onto the road from the engine of a moving car every 5 seconds. This car has been traveling for 600 meters. What is the average speed of the car?
2) A skateboarder starts from rest and moves down a hill with constant acceleration in a straight line, travelling for 6 seconds. In a second trial he starts from rest, and moves along the same straight line with the same acceleration for 2 seconds. How does his displacement from his starting point in the second trial compare with the first trial?
A) One-third as large
B) Three times larger
C) one ninth as large
D) nine times larger
E) 1/ (root square of 3) times as large?
3) A ball is thrown straight up in the air. For which situation are both the instantaneous velocity and the acceleration zero?
A) on the way up
B) at the top of the flight path
C) on the way down
D) half way up and halfway down
E) none of the above
2) A skateboarder starts from rest and moves down a hill with constant acceleration in a straight line, travelling for 6 seconds. In a second trial he starts from rest, and moves along the same straight line with the same acceleration for 2 seconds. How does his displacement from his starting point in the second trial compare with the first trial?
A) One-third as large
B) Three times larger
C) one ninth as large
D) nine times larger
E) 1/ (root square of 3) times as large?
3) A ball is thrown straight up in the air. For which situation are both the instantaneous velocity and the acceleration zero?
A) on the way up
B) at the top of the flight path
C) on the way down
D) half way up and halfway down
E) none of the above
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1) Alright then with 6 droplets (a droplet every 5 seconds) gives you that the car was in motion for 30 seconds. It covered 600 meters in that time. Average velocity to cover that distance is simply distance covered in the time it took so 600 meters/30 seconds. So it was moving with an average velocity of 20 m/s or 44.7 mph (a much more reasonable value.
It's possible that the first droplet fell right at the beginning of its movement - which would mean it only took 25 seconds. In this case it had to move at 24 m/s. I think the first answer is how the problem most likely intended it to be interpreted though.
2) Displacement of second trial is 1/9th that of the first.
dx_1= 1/2 a t^2=1/2*a*36=18*a
dx_2=1/2*a*2^2=2^a
Acceleration is stated to be the same in each case -> displacement in first case is therefore 9 times that in the second case.
3) E: none of the above
The instantaneous velocity is 0 at the peak of the flight - however it is being accelerated downward by gravity at 9.8m/s^2 still. In fact, it is accelerated constantly at a downward rate of 9.8 m/s^2 (were acceleration 0 at the peak that would imply the forces were balanced and that it was suspended there by another force in opposition to gravity - it is not.)
It's possible that the first droplet fell right at the beginning of its movement - which would mean it only took 25 seconds. In this case it had to move at 24 m/s. I think the first answer is how the problem most likely intended it to be interpreted though.
2) Displacement of second trial is 1/9th that of the first.
dx_1= 1/2 a t^2=1/2*a*36=18*a
dx_2=1/2*a*2^2=2^a
Acceleration is stated to be the same in each case -> displacement in first case is therefore 9 times that in the second case.
3) E: none of the above
The instantaneous velocity is 0 at the peak of the flight - however it is being accelerated downward by gravity at 9.8m/s^2 still. In fact, it is accelerated constantly at a downward rate of 9.8 m/s^2 (were acceleration 0 at the peak that would imply the forces were balanced and that it was suspended there by another force in opposition to gravity - it is not.)