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QA1(b):
The following is the distance-time table of an object in motion:
(b) What do you infer about the forces acting on the object?
The following is the distance-time table of an object in motion:
(b) What do you infer about the forces acting on the object?
Solution :
Step 1: Analyzing the Kinematic Data
To infer the forces acting on the object, we must first analyze the nature of the motion by calculating the speed ($v$) at different time intervals. The object's motion is defined by the provided table (Time $t$ in seconds, Distance $s$ in meters):
| Time ($t$) in s | Distance ($s$) in m |
|---|---|
| 0 | 0 |
| 2 | 1 |
| 4 | 8 |
| 6 | 27 |
| 8 | 64 |
| 10 | 125 |
Step 2: Mathematical Derivation of Motion Type
We observe that the distance traveled $s$ is proportional to the cube of time $t$. Specifically, $s = k \cdot t^3$. Let us verify the proportionality constant ($k$):
- At $t = 2, s = 1 \implies 1 = k(2^3) \implies k = 1/8$
- At $t = 4, s = 8 \implies 8 = k(4^3) \implies 8 = k(64) \implies k = 1/8$
- At $t = 10, s = 125 \implies 125 = k(10^3) \implies 125 = k(1000) \implies k = 1/8$
Since $s = \frac{1}{8} t^3$, the velocity ($v$) is the first derivative of displacement with respect to time [Per the definition of instantaneous velocity]:
$v = \frac{ds}{dt} = \frac{d}{dt}(\frac{1}{8} t^3) = \frac{3}{8} t^2$
The acceleration ($a$) is the first derivative of velocity with respect to time [Per the definition of instantaneous acceleration]:
$a = \frac{dv}{dt} = \frac{d}{dt}(\frac{3}{8} t^2) = \frac{3}{4} t$
Step 3: Inference Regarding Forces
Since the acceleration $a = 0.75t$ is not constant and increases linearly with time, we apply Newton's Second Law of Motion ($F = ma$):
Given that the mass ($m$) of the object is constant, if the acceleration is changing, the net external force ($F$) acting on the object must also be changing linearly with time [Per Newton’s Second Law of Motion: $F_{net} = m \cdot a(t)$].
Step 4: Conclusion
Because the acceleration is non-zero and non-uniform, the object is undergoing non-uniform acceleration. This implies that there is an unbalanced, time-dependent force acting upon the object. A state of constant velocity (which would imply zero net force) is not maintained, and the acceleration is not constant (which would imply a constant force). Therefore, the force must be varying to satisfy the condition of $F(t) = m \cdot (\frac{3}{4}t)$.
Final Solution: The object is moving with non-uniform acceleration; therefore, there is an unbalanced, time-varying external force acting on the object.
More Questions from Class 9 Science Force and Laws of Motion Additional Exercises
- QA1(a): The following is the distance-time table of an object in motion: (a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
- QA2: Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 $m s^{-2}$. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
- QA3: A hammer of mass 500 g, moving at 50 $m s^{-1}$, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
- QA4: A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
CBSE Solutions for Class 9 Science Force and Laws of Motion
Chapters in CBSE - Class 9 Science
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