If an object moves at a non-zero constant acceleration for a certain i

If an object moves at a non-zero constant acceleration for a certain interval of time, then the distance it covers in that time

[amp_mcq option1=”depends on its initial velocity.” option2=”is independent of its initial velocity.” option3=”increases linearly with time.” option4=”depends on its initial displacement.” correct=”option1″]

This question was previously asked in

UPSC NDA-2 – 2019

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The distance covered by an object moving with constant acceleration is given by the kinematic equation: $s = ut + \frac{1}{2}at^2$, where $s$ is the distance (or displacement magnitude if direction is constant), $u$ is the initial velocity, $a$ is the constant acceleration, and $t$ is the time interval. This equation clearly shows that the distance covered ($s$) depends on the initial velocity ($u$), assuming $a \neq 0$ and $t > 0$.

– For motion with constant acceleration, the relationship between distance, initial velocity, acceleration, and time is described by standard kinematic equations.
– The equation $s = ut + \frac{1}{2}at^2$ explicitly includes the initial velocity $u$.

– Option B is incorrect because the term $ut$ makes the distance dependent on initial velocity.
– Option C is incorrect because the presence of the $\frac{1}{2}at^2$ term (with $a \neq 0$) means the distance increases quadratically with time, not linearly.
– Option D is incorrect; the distance covered (change in position) is independent of the starting position (initial displacement). Initial displacement affects the final position, but not the distance traveled during the interval.

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