The positions of two cars A and B are given b | vedclass

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

Question

(B) Given the position functions for cars $A$ and $B$:$X_A(t) = at + bt^2$$X_B(t) = ft - t^2$The velocity of a particle is the time derivative of its

Vedclass · Accepted Answer

$\frac{f-a}{2(1+b)}$

Vedclass · Answer

(B) Given the position functions for cars $A$ and $B$:$X_A(t) = at + bt^2$$X_B(t) = ft - t^2$The velocity of a particle is the time derivative of its position function,$v = \frac{dX}{dt}$.For car $A$:$v_A = \frac{d}{dt}(at + bt^2) = a + 2bt$For car $B$:$v_B = \frac{d}{dt}(ft - t^2) = f - 2t$To find the time when both cars have the same velocity,we set $v_A = v_B$:$a + 2bt = f - 2t$Rearranging the terms to solve for $t$:$2bt + 2t = f - a$$t(2b + 2) = f - a$$t(2(b + 1)) = f - a$$t = \frac{f - a}{2(1 + b)}$Thus,the cars will have the same velocity at time $t = \frac{f - a}{2(1 + b)}$.

The positions of two cars $A$ and $B$ are given by $X_A = at + bt^2$ and $X_B = ft - t^2$. At what time will both cars have the same velocity?

Similar Questions

The velocity $v$ of a particle is given by the equation $v = 6t^2 - 6t^3$,where $v$ is in $m/s$ and $t$ is time in $s$. Then:

Draw the $x-t$ graph for positive acceleration.

Which of the following velocity-time graphs represents uniform motion?

Velocity of a particle moving along a straight line is given by $v = t^2 - 3t - 4 \, (m/s)$. Find the velocity at the instant when no net force is applicable on the particle.......$m/s$ (in $.25$)

For the velocity-time graph shown in the figure,the distance covered by the body in the last two seconds of its motion is $S_1$. What is the ratio of $S_1$ to the total distance covered by it?

Vedclass Test Series

Exam Paper Generator

Online Exam Module