Unit: 1. Kinematics
Weightage : 10-15%
An object moves along the \(x\)-axis with a velocity \(v\) given as a function of time \(t\), as shown in the graph. During which section(s) of the graph is the object’s acceleration constant, but not zero?
Correct Answer: \( \boxed{\mathrm{A}} \)
Acceleration is the rate of change of velocity with time:
\(a=\dfrac{\Delta v}{\Delta t}=\dfrac{dv}{dt}.\)
On a velocity-time graph, the acceleration is equal to the slope of the graph.
Therefore, the object’s acceleration is constant but not zero only during segments AB and BC. Hence, the correct answer is Option (A).
In an experiment, a ball is launched vertically upward. A motion sensor is placed directly below the ball as shown so that it can collect data for the height \(h\) of the ball and the velocity \(v\) of the ball as functions of time \(t\) for the trip of the ball upward and back down to its original launch position. Air resistance is negligible.
Which of the following shows variables that can be plotted such that they will create a straight line with a slope approximately equal to the acceleration of the ball?
Correct Answer: \( \boxed{\mathrm{E}} \)
For motion with constant acceleration, the velocity and time are related by the kinematic equation
\(v=v_0+at.\)
This equation has the form of a straight line,
\(y=mx+b,\)
where \(v\) is plotted on the vertical axis, \(t\) on the horizontal axis, the slope is \(a\), and the vertical intercept is \(v_0\).
Since the ball moves only under the influence of gravity, its acceleration remains constant:
\(a=-g\approx-9.8~\mathrm{m/s^2}.\)
Therefore, a graph of velocity versus time is a straight line whose slope is equal to the acceleration of the ball.
Hence, the correct answer is Option (E).
Astronauts travel to another planet and wish to determine an accurate value for the acceleration due to gravity. They drop an object from the edge of a very tall cliff and, with a motion sensor, collect the data shown below.
If air resistance is negligible, the magnitude of the acceleration of the projectile is most nearly
Correct Answer: \( \boxed{\mathrm{E}} \)
Since the object is dropped from rest, its vertical displacement is given by
\(y=v_0t+\dfrac{1}{2}at^2\)
Because \(v_0=0\),
\(y=\dfrac{1}{2}at^2\)
Thus, a graph of distance fallen versus \(t^2\) is a straight line whose slope is
\(\dfrac{1}{2}a\)
Using the data, for example at \(t=1.0\,\mathrm{s}\),
\(\dfrac{y}{t^2}=\dfrac{26.4}{1.0^2}=26.4\,\mathrm{m/s^2}\)
Therefore,
\(\dfrac{1}{2}a=26.4\,\mathrm{m/s^2}\)
\(a=2(26.4)=52.8\,\mathrm{m/s^2}\)
This value is consistent with the remaining data points, confirming that the motion has constant acceleration.
Therefore, the correct answer is (E).
