Problems Based on Graphical Representation of Motion for Class 9 Science

Answer & Solution

Answer: (i) $2.3\text{ m/s}^2$ (ii) $0\text{ m/s}^2$ (iii) $-2.3\text{ m/s}^2$

Explanation: Acceleration is the slope of the velocity-time graph $a = \frac{\Delta v}{\Delta t}$. Reading the coordinates from the graph:

(i) First 2 seconds (0s to 2s): The velocity goes from 0 to 4.6 m/s.

(ii) Between 2nd and 10th second: The velocity is constant at 4.6 m/s, so the slope is a horizontal line.

(iii) Last two seconds (10s to 12s): The velocity drops from 4.6 m/s back to 0.

Answer & Solution

Answer: No, it does not represent a real physical situation.

Explanation: When analyzing physical graphs, there are two primary reasons a distance-time graph might be impossible:
1. Time cannot go backward: If the graph curves back on itself (having two distance values for a single point in time), it implies the object exists in two places simultaneously, which is impossible.
2. Distance cannot decrease: The total path length covered by a moving object (distance) never decreases with time. If the graph shows a negative slope (going downward), it is impossible for a distance-time graph (though it would be perfectly valid for a displacement-time graph).

Answer & Solution

Answer: $25\text{ m}$

Explanation: On a velocity-time graph, a uniform velocity of 5 m/s is represented by a straight horizontal line parallel to the time axis at the 5 m/s mark.

The distance travelled is found by calculating the area under the velocity-time graph. For a time interval of 5 seconds, this shapes a rectangle with length 5 (time) and height 5 (velocity).

Answer & Solution

Answer: (a) Uniform Acceleration (b) Constant Speed (c) Uniform Retardation.

Explanation:
(a) OA: The straight line sloping upward indicates speed is increasing at a constant rate, representing uniform acceleration.
(b) AB: The horizontal line indicates speed is not changing, representing constant speed (zero acceleration).
(c) BC: The straight line sloping downward indicates speed is decreasing at a constant rate, representing uniform retardation.
(d) Acceleration: Calculated by finding the slope of OA. Read the maximum speed ($v$) at point A and the time ($t_A$). $a = \frac{v – 0}{t_A – 0}$.
(e) Retardation: Calculated by finding the magnitude of the slope of BC. Read the time at B ($t_B$) and C ($t_C$). Retardation $= \frac{v – 0}{t_C – t_B}$.
(f) Distance A to B: The area of the rectangle formed under AB. Distance $= \text{speed } v \times (t_B – t_A)$.

Answer & Solution

Explanation: The distance travelled is the area enclosed under the respective sections of the speed-time graph.

(i) Distance O to A: Find the area of the triangle formed under OA.
$\text{Area} = \frac{1}{2} \times \text{base (time interval for OA)} \times \text{height (speed at A)}$

(ii) Distance B to C: Find the area of the triangle formed under BC.
$\text{Area} = \frac{1}{2} \times \text{base (time interval for BC)} \times \text{height (speed at B)}$

(iii) Total Distance in 16 sec: Find the total area under the entire graph OABC. This can be done by summing the areas of triangle OA, rectangle AB, and triangle BC, OR by calculating the area of the entire trapezium directly:
$\text{Total Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height (max speed)}$

Answer & Solution

Explanation: Plot the given values with time on the X-axis and speed on the Y-axis. Since acceleration is uniform, you will get a straight line.

(i) Acceleration: Pick any two points on your straight line graph, $(t_1, v_1)$ and $(t_2, v_2)$. The acceleration is the slope:

(ii) Distance Travelled (50s): The distance is the area bounded by the graph and the time axis from $t=0$ to $t=50$. If the graph starts from rest ($0,0$), it’s a triangle ($\frac{1}{2} \times \text{base} \times \text{height}$). If the car had an initial speed, it forms a trapezium ($\frac{1}{2} \times \text{sum of parallel sides} \times \text{base}$).

Answer & Solution

Answer: No, it does not represent a real physical situation.

Explanation: Check the behavior of the graph line:
1. Distance cannot decrease: Distance is a scalar quantity representing the total path covered. It can only increase or remain constant (if the object stops). It can never decrease. Only displacement (vector) can decrease or become negative.
2. Infinite speed is impossible: If any part of the graph is a perfectly vertical line, it indicates a change in distance with zero change in time, implying infinite speed, which is impossible.
3. Time is unidirectional: The graph cannot form a loop or curve backward on the X-axis.

Answer & Solution

Explanation: In a position-time (or distance-time) graph, the speed is determined by calculating the slope of the line for that specific interval.

For any segment between two points $(t_1, x_1)$ and $(t_2, x_2)$, use the formula:

(i) A to B: Find the slope of segment AB using the coordinates of A and B.
(ii) B to C: If the line is horizontal, the position is not changing, meaning the slope and speed are zero.
(iii) C to D: Find the slope of segment CD. If the line goes downward, calculate the magnitude of the slope to find the speed (velocity would be negative).

Answer & Solution

Explanation: Plot the given values with time on the X-axis and speed on the Y-axis to form a straight line.

(i) Acceleration: Determine the slope of the graph.

(ii) Distance in 10s: Calculate the area under the speed-time graph from $t=0$ to $t=10$. Depending on your plotted data points, use the area of a triangle ($A = \frac{1}{2}bh$) or a trapezium ($A = \frac{1}{2}(a+b)h$) to find the total distance travelled.

Answer & Solution

Answer: (i) $40\text{ km/hr}$ (ii) $10\text{ km/hr}^2$ (iii) $0\text{ km/hr}^2$

Explanation: Acceleration is the slope of the velocity-time graph.
(i) Velocity at point C: Read directly from the y-axis at point C. The graph shows a horizontal line starting from B at $40\text{ km/hr}$, therefore the velocity at C is $40\text{ km/hr}$.
(ii) Acceleration between A and B: Point A is at $(0\text{ hr}, 20\text{ km/hr})$ and Point B is at $(2\text{ hr}, 40\text{ km/hr})$. The slope is:

(iii) Acceleration between B and C: Point B is at $40\text{ km/hr}$ and Point C is at $40\text{ km/hr}$ at 7 hours. The velocity is constant (a horizontal line), so there is no change in velocity.