A body starts from rest at x=0, t=0 with constant acceleration a . Simultaneously another body passes x=0 with constant speed v . The position of the - Physics MCQ AIEEE 2026 | ExamTest.live

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A body starts from rest at x = 0, t = 0 with constant acceleration a . Simultaneously another body passes x = 0 with constant speed v . The position of the first is x_{1} (t) and the second is x_{2} (t) . Which graph describes (x_{1} - x_{2}) vs t ?

Solution & Step-by-step Explanation

x_{1} (t) = \frac{1}{2} a t^{2} . x_{2} (t) = v t . x_{1} - x_{2} = \frac{1}{2} a t^{2} - v t .This is a quadratic equation (parabola) in t . At t = 0, y = 0 .The derivative \frac{d}{d t} (x_{1} - x_{2}) = a t - v .Initially (t < v / a), the slope is negative, so the graph goes below the t -axis before turning upwards.

Try it yourself before checking the explanation above.

A body starts from rest at x = 0, t = 0 with constant acceleration a . Simultaneously another body passes x = 0 with constant speed v . The position of the first is x_{1} (t) and the second is x_{2} (t) . Which graph describes (x_{1} - x_{2}) vs t ?

A straight line through origin

A parabola starting from origin, going negative then positive

A parabola starting from origin and staying positive

A horizontal line

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