Two cars A and B initially at rest are moving | vedclass

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

Question

(B) Using the first equation of motion, $v = u + at$. Since both cars start from rest $(u = 0)$, their velocities after time $t$ are:$v_1 = a_1 t$ and

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) Using the first equation of motion, $v = u + at$. Since both cars start from rest $(u = 0)$, their velocities after time $t$ are:$v_1 = a_1 t$ and $v_2 = a_2 t$Subtracting these, we get: $(v_1 - v_2) = (a_1 - a_2)t \dots (1)$Using the second equation of motion, $s = ut + \frac{1}{2}at^2$. The distance between them is:$s_1 - s_2 = \frac{1}{2}a_1 t^2 - \frac{1}{2}a_2 t^2 = \frac{1}{2}(a_1 - a_2)t^2$Given $s_1 - s_2 = 50 \ m$ and $(a_1 - a_2) = 4 \ m \ s^{-2}$, we have:$50 = \frac{1}{2} 	imes 4 	imes t^2 \Rightarrow 50 = 2t^2 \Rightarrow t^2 = 25 \Rightarrow t = 5 \ s$Substituting $t = 5 \ s$ and $(a_1 - a_2) = 4 \ m \ s^{-2}$ into equation $(1)$:$(v_1 - v_2) = 4 	imes 5 = 20 \ m \ s^{-1}$.

Similar Questions

$A$ body falling for $2 \, s$ covers a distance $S$ equal to that covered in the next second. Taking $g = 10 \, m/s^2$,$S = .......... m$

$A$ bullet fired into a fixed target loses half of its velocity after penetrating $3\,cm$. How much further will it penetrate before coming to rest,assuming that it faces constant resistance to motion? (in $cm$)

The initial velocity of a particle is $u$ at $t=0$ and the acceleration $a$ is given by $a = \alpha t^{3/2}$. Which of the following relations is valid?

The distance travelled by a particle is directly proportional to $t^{1/2}$,where $t =$ time elapsed. What is the nature of motion?

If average and instantaneous velocities are same,then what is the type of motion?

Vedclass Test Series

Exam Paper Generator

Online Exam Module