A car travels forward with constant velocity. It goes over a small stone, which gets stuck in the groove of a tire. The initial acceleration of the stone, as it leaves the surface of the road is,
a)vertically up
b)horizontally forward
c)horizontally backwards
d) zero
e)greater than zero but less than 45 degrees below the horizontal
I think the answer would be d)zero since the car's tire is going at constant velocity. Would my thinking be correct?
The correct answer is "vertically up". The car's tire is going at constant speed, but the velocity is constantly changing direction. The instantaneous velocity of the part of the tire that is in contact with the road is zero, so the stone is not subjected to a tangential acceleration.
From the moment it is caught in the tire it is rotating with the tire. An object that is rotating around some point with constant speed is accelerating toward that point with the centripetal acceleration of v^2/r.
A vertically upward since that is the direction of centripetal acceleration at that time.
the acceleration is upward, but it's not due to centripetal acceleration because it's velocity is zero as it stuck in the tire. As the tire moves forward, you can draw a series pictures to show that the stone is moving upward.
Ah, I see what you're getting at. You're right that the car's tire is traveling at constant velocity, but when the small stone gets stuck in the groove of the tire, it experiences a change in velocity. This change in velocity is due to the centripetal acceleration acting on the stone as it moves in a circular path with the tire.
So, the correct answer is indeed "vertically up." The stone experiences an initial acceleration that is directed upward as it leaves the surface of the road. It's kind of like the stone hitched a ride on a mini roller coaster!
Good thinking, though! Keep those gears turning!
To understand why the stone's initial acceleration is vertically up, let's break it down step by step:
1. The car is traveling forward with a constant velocity. This means that the car's speed does not change, but the direction of its velocity does not change either.
2. As the car goes over a small stone, the stone gets stuck in the groove of a tire.
3. The stone is initially in contact with the road, and as the tire moves forward, it pushes against the stone.
4. Since the stone is stuck in the groove of the tire, it moves along with the tire as it rotates.
5. However, the part of the tire that is in contact with the road has a velocity equal to the car's constant velocity. This means that the instantaneous velocity of the part of the tire in contact with the road is zero, as it is not moving relative to the road.
6. The stone, being stuck in the groove of the tire, also has an instantaneous velocity of zero at the point of contact with the road.
7. When the stone leaves the road surface, it experiences a change in its velocity. Since the stone's velocity was initially zero (as it was in contact with the road), any change in velocity would imply an acceleration.
8. Since the stone is moving upwards after leaving the road surface, its initial acceleration is vertically up. This means option a) "vertically up" is the correct answer.
In summary, even though the car's tire is moving forward at a constant velocity, the stone experiences an initial acceleration when it exits the road surface due to the change in its velocity.
A car traveling in a straight line has a velocity
of +4.4 m/s. After an acceleration of 0.65
m/s2
, the car’s velocity is +8.8 m/s.
In what time interval did the acceleration
occur?