Question:
Two boys are standing at ends $A$ and $B$ of a ground where $AB = 200\, m$. The boy at B starts running in a direction perpendicular to $AB$ with a speed of $6\, m \,s^{-1}$? The boy at A starts simultaneously with a velocity of $10 \, m \, s^{-1}$ and catches the other at time t where the time $t$ is
Two boys are standing at ends $A$ and $B$ of a ground where $AB = 200\, m$. The boy at B starts running in a direction perpendicular to $AB$ with a speed of $6\, m \,s^{-1}$? The boy at A starts simultaneously with a velocity of $10 \, m \, s^{-1}$ and catches the other at time t where the time $t$ is
Updated On: May 11, 2024
- 50 s
- 20 s
- 25 s
- 12.5 s
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The Correct Option is C
Solution and Explanation
The two boys meet at $C$ after a time $t$.
Taking horizontal motion of boy at $A$ from $A$ to $C, 200 = (10 \cos \theta)t$
or $t =\frac{200}{10 \cos \theta} = \frac{200}{10\left(\frac{\sqrt{\left(10\right)^{2} -\left(6\right)^{2}}}{10}\right)} $
$= \frac{200}{\sqrt{\left(10\right)^{2} - \left(6\right)^{2}}} = \frac{200}{8} = 25 s$
Taking horizontal motion of boy at $A$ from $A$ to $C, 200 = (10 \cos \theta)t$
or $t =\frac{200}{10 \cos \theta} = \frac{200}{10\left(\frac{\sqrt{\left(10\right)^{2} -\left(6\right)^{2}}}{10}\right)} $
$= \frac{200}{\sqrt{\left(10\right)^{2} - \left(6\right)^{2}}} = \frac{200}{8} = 25 s$
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration