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QJEEM24Ph8a38: JEE Main 2024 Physics - Class 11 JEEM24Ph8a38 Solution: A particle moving in a straight line covers half the distance…

A particle moving in a straight line covers half the distance with speed 6 m/s6 \mathrm{~m} / \mathrm{s} . The other half is covered in two equal time intervals with speeds 9 m/s9 \mathrm{~m} / \mathrm{s} and 15 m/s15 \mathrm{~m} / \mathrm{s} respectively. The average speed of the particle during the motion is :

Multiple Choice Options

  • A 9.2 m/s
  • B 8.8 m/s
  • C 10 m/s
  • D 8 m/s

Step-by-Step Verified Solution

Let's denote the total distance covered by the particle as 2d2d , where dd is the distance for each half. To calculate the average speed, we need to find the total distance traveled and divide it by the total time taken.

For the first half of the journey, the particle covers the distance dd at a speed of 6 m/s6 \, \text{m/s} . The time taken for this part of the journey can be calculated using the formula time=distancespeed\text{time} = \frac{\text{distance}}{\text{speed}} . So,

time1=d6\text{time}_1 = \frac{d}{6}

For the second half of the journey, the distance dd is further divided into two parts, each covered in equal time intervals. Given the speeds are 9 m/s9 \, \text{m/s} and 15 m/s15 \, \text{m/s} respectively, let's call the equal time intervals tt . The distances covered in these intervals can be found by distance=speed×time\text{distance} = \text{speed} \times \text{time} .

For the part covered at 9 m/s9 \, \text{m/s} :

d1=9td_1 = 9t

For the part covered at 15 m/s15 \, \text{m/s} :

d2=15td_2 = 15t

Since these two parts together make up the second half of the journey,

d1+d2=dd_1 + d_2 = d

9t+15t=d9t + 15t = d

This gives us 24t=d24t = d , and from this, we can find t=d24t = \frac{d}{24} .

The total time for the second half of the journey is the sum of the times for the two parts, which are equal ( tt each), so the total time for the second half is 2t2t . Since t=d24t = \frac{d}{24} ,

time2=2×d24=d12\text{time}_2 = 2 \times \frac{d}{24} = \frac{d}{12}

The total time taken for the entire journey is the sum of the times for the first and second halves:

total time=time1+time2=d6+d12\text{total time} = \text{time}_1 + \text{time}_2 = \frac{d}{6} + \frac{d}{12}

total time=2d12+d12=3d12=d4\text{total time} = \frac{2d}{12} + \frac{d}{12} = \frac{3d}{12} = \frac{d}{4}

The total distance is 2d2d , and the total time is d4\frac{d}{4} . Therefore, the average speed is calculated as:

average speed=total distancetotal time=2dd4=2d1×4d=8 m/s\text{average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{2d}{\frac{d}{4}} = \frac{2d}{1} \times \frac{4}{d} = 8 \, \text{m/s}

Thus, the correct answer is Option D: 8 m/s.

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