A car is moving with a speed of 150 km/h and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The stopping distance $d$ is given by the formula $d = \frac{v^2}{2a}$,where $v$ is the initial velocity and $a$ is the magnitude of deceleration.

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) The stopping distance $d$ is given by the formula $d = \frac{v^2}{2a}$,where $v$ is the initial velocity and $a$ is the magnitude of deceleration.Since the braking acceleration $a$ remains constant,the stopping distance is directly proportional to the square of the initial velocity: $d \propto v^2$.Let the initial speed be $v_1 = 150 \ km/h$ and the initial stopping distance be $d_1 = 27 \ m$.The new speed is $v_2 = \frac{1}{3} v_1$.Therefore,the new stopping distance $d_2$ is given by $d_2 = (\frac{1}{3})^2 	imes d_1 = \frac{1}{9} 	imes 27 \ m = 3 \ m$.

$A$ car is moving with a speed of $150 \ km/h$ and after applying the brakes,it travels $27 \ m$ before it stops. If the same car is moving with a speed of one-third of the initial speed,then it will stop after traveling how many meters?

Similar Questions

$A$ point moves with uniform acceleration and $v_1, v_2$ and $v_3$ denote the average velocities in the three successive intervals of time $t_1, t_2$ and $t_3$. Which of the following relations is correct?

$A$ car moving with uniform acceleration covers a distance of $200 \,m$ in the first $2 \,s$ and a distance of $220 \,m$ in the next $4 \,s$. The velocity of the car after $7 \,s$ is: (in $\,m/s$)

$A$ bullet loses $1/20$ of its velocity after passing through a wooden plank. What is the minimum number of such planks required to stop the bullet completely?

$A$ body starts from rest with uniform acceleration and its velocity at a time of $n$ seconds is $v$. The total displacement of the body in the $n^{\text{th}}$ and $(n-1)^{\text{th}}$ seconds of its motion is:

The position of a particle as a function of time $t$ is given by $x(t) = at + bt^2 - ct^3$,where $a, b,$ and $c$ are constants. When the particle attains zero acceleration,its velocity will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module