Question

Two cars aand b are at rest at the origin o. if a starts with uniform velocity 20m/s at thew same direction with constant acceleration 2m/s then at what time they will meet

Answer 1

Cars A and B start from origin. Car A travels with a uniform velocity of 20 m/s. Car B starts with a constant acceleration of 2 m/s².

To find:

Time after which they will meet

Concept:

Both the cars will meet only after covering an equal distance. So , we need to find out the distance and equate them.

Calculation:

Distance travelled by car A in time t :

$d1 = v \times t$

$= > d1 = 20 \times t$

$= > d1 = 20t$

Distance travelled by car B in time t :

$d2 = ut + \frac{1}{2} a {t}^{2}$

$= > d2 = 0 + \{ \frac{1}{2} \times 2 \times {t}^{2} \}$

$= > d2 = {t}^{2}$

Equating the 2 distances travelled :

$d1 = d2$

$= > 20t = {t}^{2}$

$= > {t}^{2} - 20t = 0$

$= > t(t - 20) = 0$

So , w can say that :

$\therefore \: \: t - 20 = 0$

$= > t = 20 \: sec$

So they will meet after 20 secs.

Answer 2

GiveN :

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 2 m/s²
• Final velocity (v) = 20 m/s

To FinD :

• At what time cars will meet

SolutioN :

Let the distance traveled by Car A is $\sf{S_1}$

And, distance traveled by Car B is $\sf{S_2}$

Both, will catch each other when they will travel equal distance.

$\underbrace{\sf{Use \: 2nd \: equation \: of \: motion}}$

$\implies \sf{S \: = \: S_1 \: = \: S_2} \\ \\ \implies \sf{S \: = \: ut \: + \: \dfrac{1}{2} at^2 } \\ \\ \implies \sf{S \: = \: 0 \: \times \: t \: + \: \dfrac{1}{\not{2}} \: \times \: \not{2} \: \times \: t^2} \\ \\ \implies \sf{S \: = \: t^2} \\ \\ \implies \sf{Displacement \: = \: t^2} \\ \\ \implies \sf{velocity \: \times \: time \: = \: t^2} \\ \\ \implies \sf{20t \: = \: t^2} \\ \\ \implies \sf{t \: = \: 20} \\ \\ \underline{\sf{\therefore \: Both \: cars \: will \: meet \: at \: 20 \: s}}$