Unit: 1. Kinematics
Weightage : 10-15%
The acceleration vector for the first three seconds of an object moving with constant acceleration is shown.
Which of the following motion diagrams best represents the velocity vectors for the first three seconds of the object’s motion?
Correct Answer: \( \boxed{\mathrm{E}} \)
The acceleration is constant and directed to the right throughout the motion. Therefore, the object’s velocity must change according to
\( \vec{a}=\dfrac{\Delta \vec{v}}{\Delta t} \).
Since the acceleration points to the right, the velocity vector must become increasingly more positive (to the right) as time passes.
Option (E) shows the object starting from rest, \(v=0\), and then acquiring a rightward velocity whose magnitude increases with time, which is exactly what constant rightward acceleration produces.
The other choices either show a constant velocity, a decreasing velocity, or a velocity pattern inconsistent with constant acceleration.
Therefore, the correct answer is (E).
The figure above shows the velocity \(v\) and acceleration \(a\) vectors for a ball thrown into the air. Assuming air resistance is negligible, which of the following could be the velocity and acceleration vectors when the ball is at the highest point in its path?
Correct Answer: \( \boxed{\mathrm{D}} \)
At the highest point of the ball’s trajectory, the ball momentarily comes to rest before reversing direction. Therefore, its instantaneous velocity is
\(v=0\).
However, gravity continues to act on the ball throughout its motion. Since air resistance is neglected, the only force acting on the ball is its weight, so the acceleration remains constant and directed downward:
\(\vec{a}=-g\,\hat{j}\).
Although the velocity becomes zero at the highest point, the acceleration does not become zero. The downward acceleration causes the ball to begin moving downward immediately after reaching its maximum height.
Therefore, the correct diagram is the one showing \(v=0\) and a downward acceleration vector, which is Option (D).
The figure shows a box on a horizontal surface. Which of the following best shows the velocity \(v\) and acceleration \(a\) vectors if the box is moving along the surface to the left with increasing speed?
Correct Answer: \( \boxed{\mathrm{A}} \)
The box is moving to the left, so its velocity vector must point to the left.
Since the box is increasing its speed, the acceleration must be in the same direction as the velocity. In general,
\( \vec{a}=\dfrac{\Delta \vec{v}}{\Delta t} \).
When an object’s speed increases, its acceleration points in the same direction as its velocity. Therefore, both the velocity and acceleration vectors point to the left.
The other options either show the acceleration opposite the direction of motion (which would indicate the object is slowing down), show zero acceleration, or show motion to the right.
Therefore, the correct choice is Option (A).
