A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

mediumMCQAIEEE-CBSE-ENG-032026Mathematics

1 mark

A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r . The value of t is given by:

Solution & Step-by-step Explanation

Let t_{1} be the time of acceleration and t_{2} be the time of retardation. t = t_{1} + t_{2} . Max velocity v_{ma x} = f t_{1} = r t_{2} ⟹ t_{2} = \frac{f}{r} t_{1} . Total time t = t_{1} + \frac{f}{r} t_{1} = t_{1} (\frac{f + r}{r}) . Distance s = \frac{1}{2} v_{ma x} t (Area of v-t graph). s = \frac{1}{2} (f t_{1}) t Substitute t_{1} = \frac{r t}{f + r} : s = \frac{1}{2} f (\frac{r t}{f + r}) t = \frac{1}{2} (\frac{f r}{f + r}) t^{2} t^{2} = \frac{2 s ( f + r )}{f r} = 2 s (\frac{1}{r} + \frac{1}{f}) t = 2 s (\frac{1}{f} + \frac{1}{r})

Try it yourself before checking the explanation above.

A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r . The value of t is given by:

2 s (\frac{1}{f} + \frac{1}{r})

\frac{2 s}{1/ f + 1/ r}

2 s (f + r)

(\frac{1}{f} + \frac{1}{r}) 2 s

Practice with timed mock tests and track your performance across Mathematics.