(A) Let the initial acceleration be $a_1$. Since the car starts from rest,$u = 0$. After $t_1 = 9 \ s$,the velocity becomes $v = a_1 t_1 = 9a_1$. Thus

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(A) Let the initial acceleration be $a_1$. Since the car starts from rest,$u = 0$. After $t_1 = 9 \ s$,the velocity becomes $v = a_1 t_1 = 9a_1$. Thus,$a_1 = v/9$.Next,the motorist maintains the speed $v$ for $t_2 = 50 \ s$.Finally,the motorist decelerates to rest. Let the deceleration be $a_2$. We are given $a_2 = 3a_1 = 3(v/9) = v/3$.Using the equation of motion $v_f = u_f - a_2 t_3$,where $v_f = 0$ (final rest),$u_f = v$ (initial speed for this phase),and $t_3$ is the time taken to stop:$0 = v - (v/3) t_3$$v = (v/3) t_3$$t_3 = 3 \ s$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Medium

$A$ motorist starting a car from rest accelerates uniformly to a speed of $v \ m/s$ in $9 \ s$. He maintains this speed for another $50 \ s$ and then applies the brakes and decelerates uniformly to rest. His deceleration is numerically equal to three times his previous acceleration. Then the time during which the deceleration takes place is .......... $s$ :-

A
$3$
B
$9$
C
$27$
D
$6$

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Class 11 Questions

Class 11

Physics Questions

Chapter

Motion in Straight Line

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Uniformly Accelerated Motion

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) s$ are $125 \ m$ and $50 \ m/s$,respectively. The distance travelled by the particle in $(t+2)^{th} s$ is . . . . . . $m$.

DifficultJEE MAIN 2024

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$A$ point starts moving in a straight line with a certain acceleration. At a time $t$ after the beginning of motion,the acceleration suddenly becomes a retardation of the same value. The time in which the point returns to the initial point is:

Difficult

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$A$ car accelerates from rest at a constant rate $\alpha$ for some time,after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $t$,then the maximum velocity acquired by the car is

DifficultIIT 1978

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$A$ body starts from rest with an acceleration of $8\,m/s^2$. What distance will it cover in the $5^{th}$ second?

Easy

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Starting from rest,the acceleration of a particle is given by $a = 2(t - 1)$. The velocity of the particle at $t = 5 \, s$ is ......... $m/s$.

MediumAIIMS 2019

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The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) s$ are $125 \ m$ and $50 \ m/s$,respectively. The distance travelled by the particle in $(t+2)^{th} s$ is . . . . . . $m$.

$A$ point starts moving in a straight line with a certain acceleration. At a time $t$ after the beginning of motion,the acceleration suddenly becomes a retardation of the same value. The time in which the point returns to the initial point is:

$A$ car accelerates from rest at a constant rate $\alpha$ for some time,after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $t$,then the maximum velocity acquired by the car is

$A$ body starts from rest with an acceleration of $8\,m/s^2$. What distance will it cover in the $5^{th}$ second?

Starting from rest,the acceleration of a particle is given by $a = 2(t - 1)$. The velocity of the particle at $t = 5 \, s$ is ......... $m/s$.

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