An automobile,travelling at 40, km/h,can be s | vedclass

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

Question

(B) Using the equation of motion $v^2 - u^2 = 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the acceleration (deceleration),

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(B) Using the equation of motion $v^2 - u^2 = 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the acceleration (deceleration),and $s$ is the stopping distance.For the first case: $u_1 = 40\, km/h$,$v_1 = 0$,$s_1 = 40\, m$.$0^2 - u_1^2 = 2a(40) \implies a = -u_1^2 / 80$.Since the deceleration $a$ is constant for the same braking force,we have $s \propto u^2$.Therefore,$s_2 / s_1 = (u_2 / u_1)^2$.Given $u_2 = 80\, km/h$,$u_1 = 40\, km/h$,and $s_1 = 40\, m$.$s_2 = 40 	imes (80 / 40)^2 = 40 	imes 2^2 = 40 	imes 4 = 160\, m$.The minimum stopping distance is $160\, m$.

An automobile,travelling at $40\, km/h$,can be stopped at a distance of $40\, m$ by applying brakes. If the same automobile is travelling at $80\, km/h$,the minimum stopping distance,in metres,is (assume no skidding)..........$m$

Similar Questions

The velocity of a bullet is reduced from $200\; m/s$ to $100\; m/s$ while travelling through a wooden block of thickness $10\; cm$. The retardation,assuming it to be uniform,will be ........... $\times 10^4\; m/s^2$.

$A$ particle moves along a straight line $OX$. At a time $t$ (in seconds),the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^3$. How long would the particle travel before coming to rest?

$A$ small block slides down on a smooth inclined plane,starting from rest at time $t=0$. Let $S_{n}$ be the distance travelled by the block in the interval $t=n-1$ to $t=n$. Then,the ratio $\frac{S_{n}}{S_{n+1}}$ is

When will the average acceleration and instantaneous acceleration of a particle be equal over any time interval?

$A$ particle moves a distance $x$ in time $t$ according to the equation $x = (t + 5)^{-1}$. The acceleration of the particle is proportional to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module