SATHEE: Kinematics Question 380

Kinematics Question 380

Question: A car accelerates from rest with a constant acceleration $α$ on a straight road. After gaining a velocity $v_{1}$ , the car moves with the same velocity for some-time. Then the car decelerated to rest with a retardation $β$ . If the total distance covered by the car is equal to S, the total time taken for it’s motion is

Options:

A) $\frac{S}{v} + \frac{v}{2} (\frac{1}{α} + \frac{1}{β})$

B) $\frac{S}{v} + \frac{v}{α} + \frac{v}{β}$

C) $(\frac{v}{α} + \frac{v}{β})$

D) $\frac{S}{v} - \frac{v}{2} (\frac{v}{α} + \frac{1}{β})$

Show Answer

Answer:

Correct Answer: A

Solution:

From v = u + at, we have $v = 0 + α t_{1} \Rightarrow \frac{v}{α}$

$0 = v - β t_{3} \Rightarrow t_{3} = \frac{v}{β} .$

$Now, S = \frac{1}{2} v t_{1} +v t_{2} + \frac{1}{2} v t_{3} = \frac{v^{2}}{2 α} + v t_{2} + \frac{v^{2}}{2 β}$
$\Rightarrow t_{2} = \frac{S}{v} - \frac{v}{2} (\frac{1}{α} + \frac{1}{β}) .$

$Hence, t_{1} + t_{2} + t_{3} = \frac{S}{v} + \frac{v}{2} (\frac{1}{α} + \frac{1}{β})$

Kinematics Question 380

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Answer:

Solution:

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Kinematics Question 380

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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