Question

Initially car Ais 10.5 m ahead of car B. Both car start moving at time t = 0 in the same direction along a straight line. The velocity-time graph of two cars is shown in figure. The time when the car B will catch car A, will be car B​

Answer 1

The distance travelled by A is calculated as SA' = 10t (velocity is given as 10m/s)

and that of B is calculated as SB'

$= \frac{1}{2 } a {t}^{2} = \frac{1}{2 } {t}^{2}$ (As, A = Tan45° = 1)

Initially A is 10.5 m ahead of B, thus we get

SA + 10.5 $= \frac{1}{2} {t}^{2}$

Thus we get the equation as

$10.5 + 10{t} = \frac{1}{2} {t}^{2}$

${t}^{2} - 20t - 21 = 0$

$t = \frac{20± \sqrt{400 + 84} }{2}$

t = 21 sec.

Hence, Both start moving at time t=0 in the same direction along a straight line. The velocity-time graph of two cars is shown in the figure. The time when the car B will catch the car A, will be 21 sec.

Hope its helpful for you.

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