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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematicsRelativeMedium3 minQB

PhysicsMediumnumerical

On a two-lane road car $A$ moves at 10 m/s; cars $B$ (behind) and $C$ (oncoming) move at 20 m/s each, both 1 km from $A$ . For $B$ to overtake $A$ before $C$ reaches $A, B$ 's minimum acceleration is:

Answer:

1.20

Solution:

$C$ reaches $A$ in
$t = \frac{1000}{20 + 10} = \frac{100}{3}$
s. Relative to $A, B$ starts 1 km behind at
$20 - 10 = 10 m / s : 1000 =$ $10 \times \frac{100}{3} + \frac{1}{2} a (\frac{100}{3})^{2}$ gives
$666.7 = \frac{10000 a}{18}$ ,
so
$a = 1.2 m / s^{2}$ .

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:Relative

⏱ 3mℹ️ Source: QB

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