A particle moving in a straight line covers half the distance with speed 6 ~m / s . The other half is covered in two equal time intervals with speeds 9 ~m / s and 15 ~m / s respectively. The average speed of the

Correct answer: B. 8 ~m / s aligned & BD S=9 t+15 t=24 t & AB S=6 t_1=24 t t_1=4 t & aligned & = dist. time = 48 t 2 t+t_1 & = 48 t 2 t+4 t 48 t 6 t 8 m / s aligned aligned

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Physics
Chapter: Motion in One Dimension

JEE Main 2024 · 9 April · Morning Shift · Physics · Motion in One Dimension

A particle moving in a straight line covers half the distance with...

A particle moving in a straight line covers half the distance with speed

6 \mathrm{~m} / \mathrm{s}

. The other half is covered in two equal time intervals with speeds

9 \mathrm{~m} / \mathrm{s}

and

15 \mathrm{~m} / \mathrm{s}

respectively. The average speed of the particle during the motion is :

Options

A $10 \mathrm{~m} / \mathrm{s}$
B $8 \mathrm{~m} / \mathrm{s}$
C $9.2 \mathrm{~m} / \mathrm{s}$
D $8.8 \mathrm{~m} / \mathrm{s}$

Correct answer

B. $8 \mathrm{~m} / \mathrm{s}$

Step-by-step solution

\begin{aligned} & \mathrm{BD} \Rightarrow \mathrm{S}=9 \mathrm{t}+15 \mathrm{t}=24 \mathrm{t} \\ & \mathrm{AB} \Rightarrow \mathrm{S}=6 \mathrm{t}_1=24 \mathrm{t} \Rightarrow \mathrm{t}_1=4 \mathrm{t} \\ & \begin{aligned} < \text{ speed }> & =\frac{\text{ dist. }}{\text{ time }}=\frac{48 \mathrm{t}}{2 \mathrm{t}+\mathrm{t}_1} \\ & =\frac{48 \mathrm{t}}{2 \mathrm{t}+4 \mathrm{t}} \Rightarrow \frac{48 \mathrm{t}}{6 \mathrm{t}} \Rightarrow 8 \mathrm{ m} / \mathrm{s}\end{aligned}

\end{aligned}$

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