The distance travelled by a body moving along a line in time $$t$$ is proportional to $$t^3$$. The acceleration-time $$(a, t)$$ graph for the motion of the body will be

Solution & Explanation

1. Relate Distance to Time Using Calculus

The problem states that the distance travelled ($$s$$) by the body is directly proportional to the cube of time ($$t^3$$). We can express this relationship mathematically by introducing a proportionality constant ($$k$$):

$$s = k \cdot t^3$$

To find the velocity ($$v$$) of the body as a function of time, we take the first derivative of distance with respect to time ($$v = \frac{ds}{dt}$$):

$$v = \frac{d}{dt}(k \cdot t^3) = 3 \cdot k \cdot t^2$$

2. Determine Acceleration as a Function of Time

To find the acceleration ($$a$$) of the body, we take the derivative of velocity with respect to time ($$a = \frac{dv}{dt}$$):

$$a = \frac{d}{dt}(3 \cdot k \cdot t^2) = 2 \cdot (3 \cdot k) \cdot t$$

$$a = 6 \cdot k \cdot t$$

Since $$6$$ and $$k$$ are constants, we can simplify this relation back into a direct proportionality statement:

$$a \propto t$$

3. Interpret the Acceleration-Time ($$a - t$$) Graph

The linear equation $$a = (6k) \cdot t$$ matches the slope-intercept form of a straight line passing through the origin ($$y = m \cdot x$$), where:

• The vertical axis variable ($$y$$) represents acceleration ($$a$$).
• The horizontal axis variable ($$x$$) represents time ($$t$$).
• The constant slope ($$m$$) is positive ($$m = 6k$$).

Therefore, the acceleration-time ($$a - t$$) graph of this body is a straight line passing through the origin with a positive slope.

Correct Option Key: A straight line passing through the origin with a positive slope.