Kinematics Long Questions

Question 11

A ball is thrown vertically upward with speed v and returns to the ground after time T. Describe the correct velocity-time graph and explain.

Answer:
Taking upward as positive, the graph is a straight line with constant negative slope -g. It begins at +v at t = 0, crosses zero at the highest point at t = T/2, and reaches -v at t = T, neglecting air resistance.

Question 12

How can velocities be found for different segments of a distance-time graph?

Answer:
For each straight segment, select two points and calculate its gradient: velocity or speed = change in distance/change in time. A rising segment has positive speed, a horizontal segment has zero speed, and a falling displacement-time segment would indicate motion in the negative direction. Convert units consistently before reporting the answer.

Question 13

Can an object's velocity be zero at an instant while its acceleration is non-zero? Give an example.

Answer:
Yes. At the highest point of a ball thrown vertically upward, its instantaneous velocity is zero, but gravitational acceleration remains approximately 10 m/s² downward. The ball therefore immediately begins to fall.

Question 14

Explain scalar and vector quantities and describe the graphical addition of vectors.

Answer:
A scalar has magnitude only, while a vector has magnitude and direction. Scalars include mass, time and energy; vectors include force, velocity and acceleration. To add vectors graphically, choose a scale and draw them head-to-tail in sequence. Draw the resultant from the tail of the first vector to the head of the last, then measure its length and direction using the selected scale and reference axes.

Question 15

Explain average speed, instantaneous speed, velocity, and uniform velocity.

Answer:
Average speed is total distance divided by total time. Instantaneous speed is speed at a specific moment. Velocity is displacement per unit time and includes direction. Uniform velocity requires constant speed along a constant direction; a change in either makes velocity non-uniform. Thus an object in circular motion may have constant speed but changing velocity.

Question 16

Describe distance-time graphs for rest, uniform speed, acceleration and deceleration.

Answer:
A horizontal line shows rest because distance is unchanged. A straight inclined line has constant gradient and shows uniform speed. An upward curve that becomes steeper shows acceleration because more distance is covered in each equal time interval. A curve whose gradient decreases shows deceleration because less distance is covered in successive equal time intervals.

Question 17

Describe speed-time graphs for constant speed and uniform acceleration, including their gradients and areas.

Answer:
A horizontal speed-time line shows constant speed and has zero gradient, so acceleration is zero. A straight rising line shows uniform acceleration; its gradient gives acceleration. The area between the graph and time axis gives distance. A rectangle gives vt for constant speed, while a triangular area gives 1/2 vt when speed rises uniformly from zero.

Question 18

A stone is dropped from a height of 45 m. Find the time taken and impact velocity using g = 10 m/s².

Answer:
Given vi = 0, S = 45 m and g = 10 m/s². From S = vit + 1/2 gt²: 45 = 5t², so t = 3 s. Then vf = vi + gt = 0 + 10(3) = 30 m/s downward.

Question 19

A motorcycle has an initial speed of 18 km/h and accelerates at 2 m/s² for 10 s. Find the distance travelled.

Answer:
Convert 18 km/h to 5 m/s. Using S = vit + 1/2 at²: S = 5(10) + 1/2(2)(10²) = 50 + 100 = 150 m.

Question 20

A ball is thrown vertically upward at 30 m/s. Find the time to reach the highest point and maximum height, using g = 10 m/s².

Answer:
Take upward as positive, so a = -10 m/s² and vf = 0 at the highest point. From vf = vi + at: 0 = 30 - 10t, giving t = 3 s. From vf² = vi² + 2aS: 0 = 900 - 20S, giving S = 45 m.