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 slope of the velocity-time graph for motion with uniform velocity is equal to:

The position $x$ of a particle moving in one dimension under the influence of a constant force is given by $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. Find the displacement of the particle in $m$ when its velocity becomes zero.

The position-time graph for a particle is shown in the figure. Starting from $t=0$,at what time $t$ (in seconds) is the average velocity zero?

The $v-t$ graphs of two moving objects make angles of $30^{\circ}$ and $45^{\circ}$ with the time axis. Find the ratio of their accelerations.

The displacement-time graphs of two moving particles make angles of $30^{\circ}$ and $45^{\circ}$ with the time-axis as shown in the figure. The ratio of their respective velocities is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module