Question

A car starting from rest and moving with uniform acceleration on a straight road travels for some time such that the path described by it in the last second and in the penultimate second of its motion are in the ratio of 4:3 respectively . Calculate the total time traveled by the body.

PLZ ANSWER

Answer 1

Solution :

⏭ Given:

✏ Initial velocity of car = zero

✏ Ratio of distance covered by car in the last second and the penultimate second is 4:3

⏭ To Find:

✏Total time traveled by the body

⏭ Formula:

✏ Formula of distance covered by moving body in nth second is given by

$\bigstar \: \boxed{\sf{\pink{d = u+\dfrac{a}{2}(2n-1)}}}$

⏭ Terms indication:

• d denotes distance
• u denotes initial velocity
• a denotes acceleration
• n denotes no. of nth second

⏭ Calculation:

$\implies\sf \: \dfrac{d1}{d2} = \dfrac{2n-1}{2(n-1)-1}\\ \\ \implies\sf \: \dfrac{4}{3} = \dfrac{2n-1}{2n-3}\\ \\ \implies\sf \: 4(2n-3) = 3(2n-1)\\ \\ \implies\sf \: 8n-12 = 6n-3\\ \\ \implies\sf \: 2n = 9\\ \\ \implies\: \boxed{\sf{\orange{n = 4.5\: sec}}}$

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✒ Total time traveled by body = 4.5sec