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Standard XII

Physics

1st Equation of Motion

Two cars are ...

Question

Two cars are travelling towards each other on a straight road at velocities $15 m s^{- 1}$ and $16 m s^{- 1}$ respectively. When they are $150$ m apart, both the drivers apply the breaks and the cars decelerates at $3 m s^{- 2}$ and $4 m s^{- 2}$ until they stop. How for apart will they be when they have come to a stop.

86.5

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89.5

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85.5

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81.36

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Solution

The correct option is D $81.36$ m
solving with respect to the car $1$ moving with speed $15 m / s$ .
using concept of relative motion we get $V_{21} = V_{2 g} - V_{1 g}$
same with acceleration $a_{21} = a_{2 g} - a_{1 g}$
so velocity car $2$ is $15 + 16 = 31 m / s, a c c = 4 + 3 = 7 m / s^{2}$ and $s e p a r a t i o n = 150 m$
distance covered before coming to rest $\frac{31^{2}}{2 \times 7} = S => S = 68.64$
so now separation between them is $150 - 68.64 = 81.36 m$
so none of the option is correct.

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Two cars are travelling towards each other on a straight road at velocities 15ms−1 and 16ms−1 respectively. When they are 150m apart, both the drivers apply the breaks and the cars decelerates at 3ms−2 and 4ms−2 until they stop. How for apart will they be when they have come to a stop.