1. A runner starts at rest and accelerates uniformly at a rate of 0.85 m/s2
. What is the runner’s
velocity after the runner has traveled 6.0 m?
A. 2.3 m/s
B. 3.2 m/s
C. 5.1 m/s
D. 10.2 m/s
2. An official drag race in the United States is usually 1/4 mile long, which is 402 m. If a car
undergoes constant acceleration and requires 9.50 s to cross the finish line, what is the average
acceleration of the car?
A. 4.45 m/s2
B. 8.91 m/s2
C. 42.3 m/s2
D. 84.6 m/s2
3. Suppose you throw a ball straight upward. Ignoring air resistance, from the instant the ball
leaves your hand, its
A. velocity increases and then decreases.
B. acceleration decreases steadily.
C. acceleration is constant and its velocity decreases.
D. acceleration is constant and its velocity increases.
4. A student uses a model catapult to launch an apple with an initial upward velocity of 52 m/s.
Ignoring the effects of air resistance, what is the velocity of the apple after 4.5 s?
A. 52 m/s
B. 96 m/s
C. 7.9 m/s
D. -44 m/s
yolo
1. To find the runner's velocity after traveling 6.0 m, we can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity (which is 0 m/s since the runner starts at rest), a is the acceleration (0.85 m/s^2), and s is the distance traveled (6.0 m).
Plugging in the values, we have:
v^2 = 0^2 + 2(0.85)(6.0)
v^2 = 0 + 10.2
v = sqrt(10.2)
v ≈ 3.2 m/s
Therefore, the runner's velocity after traveling 6.0 m is approximately 3.2 m/s. So the correct answer is B. 3.2 m/s.
2. We can use the equation of motion:
s = ut + (1/2)at^2
where s is the distance (402 m), u is the initial velocity (which we don't know), a is the acceleration, and t is the time taken (9.50 s). We are trying to find the acceleration, so we can rearrange the equation as follows:
a = 2(s - ut) / t^2
Plugging in the values, we have:
a = 2(402 - u * 9.50) / (9.50^2)
a = (804 - 9.50u) / 90.25
Since the car undergoes constant acceleration, the average acceleration is equal to the acceleration throughout the entire race. Therefore, the average acceleration of the car is (804 - 9.50u) / 90.25. However, without knowing the value of u, we cannot determine the exact average acceleration. So the correct answer is none of the given options.
3. When the ball is thrown straight upwards, it experiences a downward acceleration due to gravity. The velocity of the ball initially increases as it moves away from the ground, then it reaches its maximum height and starts to fall back down. At the very top, the velocity becomes zero, and then it starts to decrease in the opposite direction.
Therefore, the answer is A. The velocity increases and then decreases.
4. Assuming we are considering the direction of the initial upward velocity as positive, the acceleration due to gravity will be acting in the opposite direction with a magnitude of 9.8 m/s^2.
Using the kinematic equation:
v = u + at
where v is the final velocity, u is the initial velocity (52 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time (4.5 s).
Plugging in the values, we have:
v = 52 - 9.8 * 4.5
v = 52 - 44.1
v = 7.9 m/s
Therefore, the velocity of the apple after 4.5 s is 7.9 m/s. So the correct answer is C. 7.9 m/s.
1. To find the runner's velocity, we can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity (in this case, 0 m/s), a is the acceleration (0.85 m/s^2), and t is the time (unknown).
Since the runner starts at rest, the initial velocity is 0 m/s, and the equation simplifies to:
v = 0 + (0.85 m/s^2)(t)
We are given that the runner has traveled 6.0 m, so we can use this information to solve for t:
s = ut + (1/2)at^2
6.0 m = 0 + (1/2)(0.85 m/s^2)(t^2)
Rearranging the equation:
0.85 m/s^2 * t^2 = (2 * 6.0 m)
t^2 = (2 * 6.0 m) / 0.85 m/s^2
t^2 = 14.12 s^2
Taking the square root of both sides:
t = √14.12 s
Now, we substitute this value of t back into the original equation to find the final velocity:
v = 0 + (0.85 m/s^2)(√14.12 s)
v ≈ 3.2 m/s
Therefore, the runner's velocity after traveling 6.0 m is approximately 3.2 m/s. Answer choice B is correct.
2. To find the average acceleration of the car, we can use the equation of motion:
s = ut + (1/2)at^2
where s is the distance traveled (402 m), u is the initial velocity (unknown), t is the time taken (9.50 s), and a is the average acceleration (unknown).
Rearranging the equation:
a = 2(s - ut) / t^2
a = 2(402 m - u * 9.50 s) / (9.50 s)^2
Since the car undergoes constant acceleration, the final velocity can be given as:
v = u + at
Using this equation, we can eliminate u from the previous equation:
a = 2(402 m - (u + at) * 9.50 s) / (9.50 s)^2
Expanding and rearranging the equation:
38.19a = 402 m - 9.50u - 9.50^2at
Simplifying:
38.19a + 9.50^2at = 402 m - 9.50u
We now use the fact that the car crosses the finish line, so the final velocity, v, is equal to 0:
0 = u + at
Rearranging the equation:
u = -at
Substituting this into the previous equation:
38.19a + 9.50^2(-at) = 402 m - 9.50u
38.19a - 9.50^2at = 402 m + 9.50^2at
Simplifying:
(38.19 - 9.50^2)a = 402 m
a = 402 m / (38.19 - 9.50^2)
a ≈ 4.45 m/s^2
Therefore, the average acceleration of the car is approximately 4.45 m/s^2. Answer choice A is correct.
3. When a ball is thrown straight upward, its velocity and acceleration change due to the force of gravity acting on it. However, the acceleration remains constant throughout the ball's motion.
The acceleration is solely due to the force of gravity, and it acts in the downward direction. This acceleration is approximately 9.8 m/s^2 (assuming no air resistance).
As the ball moves upward, its velocity decreases because the acceleration is directed opposite to the motion. At the highest point of its trajectory, the ball briefly reaches a velocity of 0 m/s before it starts to fall downward.
Therefore, the correct answer is C. The acceleration is constant, and the velocity of the ball decreases.
4. To find the velocity of the apple after 4.5 s, we can use the equation of motion:
v = u + at
where v is the final velocity (unknown), u is the initial velocity (52 m/s), a is the acceleration (assuming constant and equal to the acceleration due to gravity, approximately -9.8 m/s^2), and t is the time (4.5 s).
Substituting the values into the equation:
v = 52 m/s + (-9.8 m/s^2)(4.5 s)
v = 52 m/s - 44.1 m/s
v ≈ 7.9 m/s
Therefore, the velocity of the apple after 4.5 s is approximately 7.9 m/s. Answer choice C is correct.