The positions of two cars A and B are X_A = a | vedclass

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Question

Explanation$:$

Given that,

Position of $\mathrm{A}=X_{a}(t)=a t+b t^{2}$

Position of $\mathrm{B}=X_{b}(t)=f t+f t^{2}$

We know that,

Ve

Vedclass · Accepted Answer

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

Vedclass · Answer

Explanation$:$

Given that,

Position of $\mathrm{A}=X_{a}(t)=a t+b t^{2}$

Position of $\mathrm{B}=X_{b}(t)=f t+f t^{2}$

We know that,

Velocity of $\mathrm{A}=v_{A}=\frac{d\left(X_{P}(t)\right)}{d t}$

$v_{A}=a+2 b t$

Velocity of $B=v_{B}=\frac{d\left(X_{Q}(t)\right)}{d t}$

$v_{B}=f+2 f t$

When the cars have same velocity,

$v_{A}=v_{B}$

$a+2 b t=f+f t$

The time will be

$t=\frac{f-a}{2(1+b)}$

Hence, At this time the cars will have same velocity.

The positions of two cars $A$ and $B$ are $X_A = at + bt^2,$ $X_B = ft -t^2$ At what time Both cars will have same velocity

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