Class 9 – Science Extra Questions – Ch. 7 – Motion

Ans: C

Explanation: Distance traveled is the total length of the path covered, regardless of direction. The person walks 5 meters east, covering a distance of 5 meters. Then, the person walks 5 meters west, covering another distance of 5 meters. The total distance traveled is the sum of these two distances: 5 meters + 5 meters = 10 meters.
Correct Option: C

Ans: B

Explanation: The distances covered by the object in successive seconds are 4.9 m, 14.7 m, and 24.5 m. Let’s find the difference between these distances:
14.7 m – 4.9 m = 9.8 m
24.5 m – 14.7 m = 9.8 m
The difference in the distances covered in successive equal time intervals is constant (9.8 m). This means the object is covering more distance in each subsequent second, and the increase in distance covered per second is uniform. This is characteristic of an object moving with uniform acceleration. Uniform motion (A) implies constant velocity and therefore equal distances covered in equal time intervals. No acceleration (C) also implies constant velocity. Uniform velocity (D) is synonymous with uniform motion. Therefore, the pattern indicates uniform acceleration. The value 9.8 m/s² is approximately the acceleration due to gravity.

Correct Option: B

Ans: C

Explanation: Uniform deceleration means that the acceleration is constant and negative. In a speed-time graph, acceleration is represented by the slope of the line. A negative slope indicates deceleration. A constant negative slope indicates uniform deceleration. Segment AB shows a constant positive slope, representing uniform acceleration. Segment BC shows a zero slope, representing constant speed (zero acceleration). Segment CD shows a constant negative slope, representing uniform deceleration. Segment DE shows an increasing positive slope, representing increasing acceleration.
Correct Option: C

Ans: A

Explanation: The question states that the car accelerates uniformly from rest. This means the initial velocity (\(v_i\)) is 0 m/s. The car reaches a final velocity (\(v_f\)) of 20 m/s in a time interval (\(\Delta t\)) of 5 seconds. Acceleration is defined as the rate of change of velocity, which can be calculated using the formula: \(a = \frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f – v_i\). Plugging in the given values, we have \(a = \frac{20 \text{ m/s} – 0 \text{ m/s}}{5 \text{ s}} = \frac{20 \text{ m/s}}{5 \text{ s}} = 4 \text{ m/s}^2\).
Correct Option: A

Ans: D

Explanation: The velocity-time graph is a straight line passing through the origin, which indicates uniform acceleration. The slope of a velocity-time graph is defined as the change in velocity divided by the change in time (Δv/Δt). This quantity is the definition of acceleration. Distance traveled and displacement are related to the area under the velocity-time graph, not its slope. Time elapsed is represented on the x-axis.
Correct Option: D

Ans: D

Explanation: Speed is the magnitude of velocity. Velocity is a vector quantity that includes both magnitude (speed) and direction, while speed is a scalar quantity that only indicates the magnitude of motion. Therefore, the speed of a moving body is equal to the magnitude of its velocity. Option C incorrectly states that speed is velocity in a given direction, as velocity is the one that has direction. Option D correctly states that velocity is speed in a given direction. Options A and B are incorrect because speed can be equal to velocity (if direction is not considered as a separate aspect from magnitude) or the magnitude of velocity, but not always higher or lower. However, the question asks for the relationship. Velocity has both magnitude and direction. Speed is the magnitude of velocity. Thus, the magnitude of velocity is the speed. Therefore, velocity is the speed in a given direction.
Correct Option: D

Ans: B

Explanation: The car starts from rest, meaning its initial velocity (\(u\)) is 0 m/s. The acceleration (\(a\)) is given as 4 m/s², and the time (\(t\)) is 10 seconds. We need to find the distance traveled (\(s\)). We can use the kinematic equation \(s = ut + \frac{1}{2}at^2\). Plugging in the values, we get \(s = (0 \text{ m/s})(10 \text{ s}) + \frac{1}{2}(4 \text{ m/s}^2)(10 \text{ s})^2\). This simplifies to \(s = 0 + \frac{1}{2}(4 \text{ m/s}^2)(100 \text{ s}^2) = 2 \text{ m/s}^2 \times 100 \text{ s}^2 = 200 \text{ m}\).
Correct Option: B

Ans: D

Explanation: First, we need to convert the velocities from km/h to m/s.
Initial velocity (u) = 20 km/h. To convert km/h to m/s, we multiply by (1000 m / 3600 s) or 5/18.
u = 20 * (5/18) m/s = 100/18 m/s = 50/9 m/s.
Final velocity (v) = 50 km/h.
v = 50 * (5/18) m/s = 250/18 m/s = 125/9 m/s.
The time taken (t) is 10 seconds.
Acceleration (a) is given by the formula: a = (v – u) / t.
a = ((125/9) m/s – (50/9) m/s) / 10 s
a = ((125 – 50) / 9) m/s / 10 s
a = (75 / 9) m/s / 10 s
a = (75 / 90) m/s²
a = (5/6) m/s²
Now we convert the fraction to a decimal: 5/6 ≈ 0.8333 m/s².

Comparing this value to the options:
A. 1 m/s2
B. 1.5 m/s2
C. 2 m/s2
D. 0.83 m/s2

The calculated acceleration is approximately 0.83 m/s², which matches option D.
Correct Option: D

Ans: B

Explanation: This question can be answered using the principles of kinematics, specifically the equation of motion relating final velocity, initial velocity, acceleration, and displacement.
The initial velocity (u) of the ball is 0 since it is dropped.
The acceleration (a) is due to gravity, which is ‘g’.
The displacement (s) is the height from which the ball is dropped, which is ‘H’.
We want to find the final velocity (v) just before it hits the ground.
The relevant kinematic equation is \(v^2 = u^2 + 2as\).
Substituting the known values:
\(v^2 = 0^2 + 2(g)(H)\)
\(v^2 = 2gH\)
Taking the square root of both sides to find v:
\(v = \sqrt{2gH}\)
Correct Option: B

Ans: A

Explanation:The formula for distance covered under uniform acceleration is \(s = ut + \frac{1}{2}at^2\), where \(s\) is the distance, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.
The object starts from rest, so the initial velocity \(u = 0\).
The distance covered in the first 5 seconds is \(D_1\). So, \(D_1 = 0 \times 5 + \frac{1}{2}a(5^2) = \frac{25}{2}a\).
The distance covered in the first 6 seconds is \(D_2\). So, \(D_2 = 0 \times 6 + \frac{1}{2}a(6^2) = \frac{36}{2}a = 18a\).
The distance covered in the 6th second is the difference between the distance covered in the first 6 seconds and the distance covered in the first 5 seconds.
Distance covered in the 6th second = \(D_2 – D_1\).
Substituting the expressions for \(D_1\) and \(D_2\):
Distance covered in the 6th second = \(18a – \frac{25}{2}a = (\frac{36}{2} – \frac{25}{2})a = \frac{11}{2}a\).
Let’s check the options.
Option A: \(D_2 – D_1 = 18a – \frac{25}{2}a = \frac{36a – 25a}{2} = \frac{11}{2}a\). This matches our result.
Option B: \(2D_1 – D_2 = 2(\frac{25}{2}a) – 18a = 25a – 18a = 7a\). This is incorrect.
Option C: \(D_1 – 2D_2 = \frac{25}{2}a – 2(18a) = \frac{25}{2}a – 36a = (\frac{25 – 72}{2})a = -\frac{47}{2}a\). Distance cannot be negative, and this does not match.
Option D: \(D_1 + D_2 = \frac{25}{2}a + 18a = (\frac{25 + 36}{2})a = \frac{61}{2}a\). This is incorrect.
Therefore, the distance covered by the object in the 6th second is \(D_2 – D_1\).
Correct Option: A