SATHEE: Kinematics Question 371

Kinematics Question 371

Question: A car, starting from rest, accelerates at the rate through a distance S, then continues at constant speed for time t and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed it’s 15 S, then

Options:

A) $S= \frac{1}{6} f t^{2}$

B) $S=ft$

C) $S= \frac{1}{4} f t^{2}$

D) $S= \frac{1}{72} f t^{2}$

Show Answer

Answer:

Correct Answer: D

Solution:

distance from A to B = $S = \frac{1}{2} f t_{1}^{}$ distance from B to C = $(f t_{1}) t$ distance from C to D = $\frac{u^{2}}{2 a} = \frac{{(f t_{1})}^{2}}{2 (f / 2)}$

$f t_{1}^{2} = 2 S$

$\Rightarrow S + f t_{1} t + 2 S = 15 S$

$\Rightarrow f t_{1} t + 12 S \dots . . (i)$

$\frac{1}{2} f t_{1}^{2} = S$ Dividing (i) by (ii), we get $t_{1} = \frac{t}{6}$

$\Rightarrow S = \frac{1}{2} f {(\frac{t}{6})}^{2} = \frac{f t^{2}}{72}$

Kinematics Question 371

Question: A car, starting from rest, accelerates at the rate through a distance S, then continues at constant speed for time t and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed it’s 15 S, then

Options:

Answer:

Solution:

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Kinematics Question 371

Question: A car, starting from rest, accelerates at the rate through a distance S, then continues at constant speed for time t and then decelerates at the rate f2 to come to rest. If the total distance traversed it’s 15 S, then

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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Question: A car, starting from rest, accelerates at the rate through a distance S, then continues at constant speed for time t and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed it’s 15 S, then