(A) Given deceleration $\omega = -\frac{dv}{dt} = a \sqrt{v}$.$1$. To find time $t$:$\frac{dv}{dt} = -a \sqrt{v} \implies \int_{v_0}^{0} v^{-1/2} dv =

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$\frac{2 v_0^{3/2}}{3 a}, \frac{2 \sqrt{v_0}}{a}$

(A) Given deceleration $\omega = -\frac{dv}{dt} = a \sqrt{v}$.$1$. To find time $t$:$\frac{dv}{dt} = -a \sqrt{v} \implies \int_{v_0}^{0} v^{-1/2} dv = \int_{0}^{t} -a dt$$[2 \sqrt{v}]_{v_0}^{0} = -at \implies -2 \sqrt{v_0} = -at \implies t = \frac{2 \sqrt{v_0}}{a}$.$2$. To find distance $s$:Using $\omega = v \frac{dv}{ds} = a \sqrt{v} \implies v \frac{dv}{ds} = -a \sqrt{v} \implies \sqrt{v} dv = -a ds$.Integrating both sides: $\int_{v_0}^{0} v^{1/2} dv = \int_{0}^{s} -a ds$$[\frac{2}{3} v^{3/2}]_{v_0}^{0} = -as \implies -\frac{2}{3} v_0^{3/2} = -as \implies s = \frac{2 v_0^{3/2}}{3 a}$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

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The deceleration of a car traveling on a straight highway is a function of its instantaneous velocity $v$ given by $\omega = a \sqrt{v}$, where $a$ is a constant. If the initial velocity of the car is $v_0$, the distance the car will travel and the time it takes before it stops are:

TS EAMCET 2017 All TS EAMCET

A
$\frac{2 v_0^{3/2}}{3 a}, \frac{2 \sqrt{v_0}}{a}$
B
$\frac{3 v_0^{3/2}}{2 a}, \frac{1}{2 a} \sqrt{v_0}$
C
$\frac{3 a}{2} v_0^{3/2}, \frac{a}{2} \sqrt{v_0}$
D
$\frac{2}{3 a} v_0, \frac{2}{a} v_0$

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Motion in Straight Line

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Uniformly Accelerated Motion

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$A$ body $A$ starts from rest with an acceleration $a_{1}$. After two seconds,another body $B$ starts from rest with an acceleration $a_{2}$. If they travel equal distances in the fifth second after the start of $A$,the ratio $a_{1}: a_{2}$ will be equal to:

MediumAIIMS 2019

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Velocity of a body moving along a straight line with uniform acceleration $a$ reduces by $\frac{3}{4}$ of its initial velocity in time $t_0$. The total time of motion of the body till its velocity becomes zero is

Medium

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Stopping distance of vehicles: When brakes are applied to a moving vehicle,the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity,or deceleration,$-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.

Easy

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Consider that a truck is moving initially with $54 \,km/h$. It is stopped by the driver after looking at an obstacle with a deceleration of $10 \,m/s^2$. The distance travelled by the truck before coming to rest is (in $\,m$)

EasyTS EAMCET 2020

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The relation between position $(x)$ and time $(t)$ is given below for a particle moving along a straight line. Which of the following equations represents uniformly accelerated motion? [where $\alpha$ and $\beta$ are positive constants]

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The deceleration of a car traveling on a straight highway is a function of its instantaneous velocity $v$ given by $\omega = a \sqrt{v}$, where $a$ is a constant. If the initial velocity of the car is $v_0$, the distance the car will travel and the time it takes before it stops are:

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$A$ body $A$ starts from rest with an acceleration $a_{1}$. After two seconds,another body $B$ starts from rest with an acceleration $a_{2}$. If they travel equal distances in the fifth second after the start of $A$,the ratio $a_{1}: a_{2}$ will be equal to:

Velocity of a body moving along a straight line with uniform acceleration $a$ reduces by $\frac{3}{4}$ of its initial velocity in time $t_0$. The total time of motion of the body till its velocity becomes zero is

Consider that a truck is moving initially with $54 \,km/h$. It is stopped by the driver after looking at an obstacle with a deceleration of $10 \,m/s^2$. The distance travelled by the truck before coming to rest is (in $\,m$)

The relation between position $(x)$ and time $(t)$ is given below for a particle moving along a straight line. Which of the following equations represents uniformly accelerated motion? [where $\alpha$ and $\beta$ are positive constants]

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