Question 14: Motion in 1D

Question

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 :

Accepted Answer

Let's denote the total distance covered by the particle as $$2d$$, where $$d$$ 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 $$d$$ at a speed of $$6 \, 	ext{m/s}$$. The time taken for this part of the journey can be calculated using the formula $$	ext{time} = \frac{	ext{distance}}{	ext{speed}}$$. So,
$$
	ext{time}_1 = \frac{d}{6}
$$
For the second half of the journey, the distance $$d$$ is further divided into two parts, each covered in equal time intervals. Given the speeds are $$9 \, 	ext{m/s}$$ and $$15 \, 	ext{m/s}$$ respectively, let's call the equal time intervals $$t$$. The distances covered in these intervals can be found by $$	ext{distance} = 	ext{speed} 	imes 	ext{time}$$.
For the part covered at $$9 \, 	ext{m/s}$$:
$$
d_1 = 9t
$$
For the part covered at $$15 \, 	ext{m/s}$$:
$$
d_2 = 15t
$$
Since these two parts together make up the second half of the journey,
$$
d_1 + d_2 = d
$$
$$
9t + 15t = d
$$
This gives us $$24t = d$$, and from this, we can find $$t = \frac{d}{24}$$.
The total time for the second half of the journey is the sum of the times for the two parts, which are equal ($$t$$ each), so the total time for the second half is $$2t$$. Since $$t = \frac{d}{24}$$,
$$
	ext{time}_2 = 2 	imes \frac{d}{24} = \frac{d}{12}
$$
The total time taken for the entire journey is the sum of the times for the first and second halves:
$$
	ext{total time} = 	ext{time}_1 + 	ext{time}_2 = \frac{d}{6} + \frac{d}{12}
$$
$$
	ext{total time} = \frac{2d}{12} + \frac{d}{12} = \frac{3d}{12} = \frac{d}{4}
$$
The total distance is $$2d$$, and the total time is $$\frac{d}{4}$$. Therefore, the average speed is calculated as:
$$
	ext{average speed} = \frac{	ext{total distance}}{	ext{total time}} = \frac{2d}{\frac{d}{4}} = \frac{2d}{1} 	imes \frac{4}{d} = 8 \, 	ext{m/s}
$$
Thus, the correct answer is
Option D: 8 m/s
.

Answer

9.2 m/s

Answer

8.8 m/s

Answer

10 m/s

Answer

8 m/s

Motion in 1D

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 :

Choose the correct answer: