Question

Consider the following questions and a student's response to each question (shown in italics). In each case, determine if the student's answer is true or false.

Explain the reasons that the distance curve has the shape it does.

The distance curve is an increasing curved line because the object is accelerating.

TrueFalse    

The velocity curve is an increasing straight line. What does this mean?

The velocity curve is a straight line because the object is moving with constant speed.

TrueFalse    

Calculate the slope of the velocity curve. Do you recognize what this is?

The slope of the velocity curve is the acceleration of the object and I calculated a slope of about 4.9 m/s².

TrueFalse    

The acceleration curve is a horizontal straight line. What does this mean?

The acceleration curve is a horizontal line because the acceleration of the object does not change with time.

TrueFalse

Q2. A ball is released from the top of a cliff; it hits the ground 3.5 seconds later. Assuming the ball is a free falling object, how many meters is the cliff above the ground below? (Use

g = 9.8 m/s².

Express your answer to one decimal point.)

Q3. The velocity vs. time curve of an object is an increasing straight line. You observe that when the time is 1 second, the velocity is equal to 4 m/s, and when the time is 4 seconds, the velocity is 28 m/s.What is the acceleration of the object? (Express your answer to one decimal point.)

Q4. What is the magnitude of the acceleration of a modified Atwood machine if the mass of the cart is 8 kg and the hanging mass is 10 kg? (Use

g = 9.8 m/s².

Express your answer to one decimal point.

Q5 In a completely frictionless Newton's Cradle toy, the change in speed of the balls coming in and those going out is zero. This is because the net force at the time of the collisions is equal to the weight of the number of balls interacting.

TrueFalse

Answer 1

The distance curve is an increasing curved line because the object is accelerating.

True it shows variable acceleration

The velocity curve is an increasing straight line. What does this mean?

The velocity curve is a straight line because the object is moving with constant speed.

False

(Because each velocity point actually represents the slope at each of the distance points for that given time. If you find the slopes of all the points on the distance-time graph, and plot the slopes against those times, you'll get a straight line.

Common sense explanation. Because, the distance-time is curved, then that means that you're either going faster or slower (depending on which way it's curved). Therefore, the velocity is either increasing or decreasing.

Rarely, you get a curved line for velocity. It's straight this time, because your acceleration is constant. But how often does acceleration change? Very very rarely. That's why your velocity curve will almost always be increasing or decreasing at the same rate, which makes it a straight line.)

Calculate the slope of the velocity curve. Do you recognize what this is?

The slope of the velocity curve is the acceleration of the object and I calculated a slope of about 4.9 m/s².

The slope of the velocity curve is the acceleration of the object is true but the statements "I calculated a slope of about 4.9 m/s²." is false

()
The acceleration curve is a horizontal straight line. What does this mean?

The acceleration curve is a horizontal line because the acceleration of the object does not change with time.

True same explanation as above

Q2.

Let Vo = initial velocity = 0 (m/s) x = vertical displacement t = time and a = gravitational constant = 9.8 meters per second squared (m/s2)
x= initial velocity * time + 1/2 acceleration * time squared = Vot + (1/2)at2 = (0*3.5) + (1/2)(9.8)(3.5)(3.5) = 60.025 meters (metric) = 60.0 meters

Q3.

Q4 use the formula a=mg/M+m = 10x9.8/(10+18) to equal 5.44 m/s^2.

Q5 FALSE

( you can find the velocities of both objects after an elastic collision
even if you don't know the velocity of either object after the collision. To do this you have to writeequations for momentum and energy conservation and solve them simultaneously. This is why the