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Question
JEE Main
Physics

Two trucks A and B are approaching each other on a straight path with the velocities of $16 \ m/s$ and $20 \ m/s$ respectively. When they are $200 \ m$ apart their drivers see each other and applies the breaks simultaneously. If the truck A decelerates with $2 \ m/s^2$ and truck B decelerates with $4 \ m/s^2$, what is the distance between them when they finally stop?

The correct answer is
$86 \ m$

Calculating Stopping Distance for Truck A

We need to find the distance Truck A travels before stopping. We use the kinematic equation $v^2 = u^2 + 2as$, where:

  • $v$ is the final velocity ($0 \ m/s$ as it stops)
  • $u$ is the initial velocity ($16 \ m/s$)
  • $a$ is the acceleration (deceleration, so $-2 \ m/s^2$)
  • $s$ is the distance traveled

Plugging in the values:

$0^2 = (16 \ m/s)^2 + 2(-2 \ m/s^2)s_A$

$0 = 256 \ m^2/s^2 - (4 \ m/s^2)s_A$

$s_A = \frac{256 \ m^2/s^2}{4 \ m/s^2} = 64 \ m$

Truck A travels $64 \ m$ before stopping.

Calculating Stopping Distance for Truck B

Similarly, we find the distance Truck B travels before stopping using the same kinematic equation $v^2 = u^2 + 2as$. For Truck B:

  • $v = 0 \ m/s$
  • $u = 20 \ m/s$
  • $a = -4 \ m/s^2$

Substituting the values:

$0^2 = (20 \ m/s)^2 + 2(-4 \ m/s^2)s_B$

$0 = 400 \ m^2/s^2 - (8 \ m/s^2)s_B$

$s_B = \frac{400 \ m^2/s^2}{8 \ m/s^2} = 50 \ m$

Truck B travels $50 \ m$ before stopping.

Determining Final Distance Between Trucks

The initial distance between the trucks was $200 \ m$. When they stop, the total distance covered by both trucks is the sum of their stopping distances.

Total distance covered $= s_A + s_B = 64 \ m + 50 \ m = 114 \ m$.

The final distance between them is the initial distance minus the total distance they covered while braking.

Final distance $= \text{Initial Distance} - (s_A + s_B)$

Final distance $= 200 \ m - 114 \ m = 86 \ m$.

Therefore, the distance between the trucks when they stop is $86 \ m$.

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