Question

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 particle during the motion is:

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

Answer 1

Answer: (D)

8 m/s

Answer 2

Step 1: Analyze the motion

• Let the total distance covered by the particle be $2S$.
• For the first half of the distance ($S$):
• For the second half of the distance ($S$), it is covered in two equal time intervals ($t, t$):
  • In the first interval ($t$), speed = $9 \, \text{m/s}$, so: $S_1 = 9t \implies t = \frac{S_1}{9}.$
  • In the second interval ($t$), speed = $15 \, \text{m/s}$, so: $S_2 = 15t.$
• Since $S_1 + S_2 = S$, solve for $t$: $9t + 15t = S \implies t = \frac{S}{24}.$

Step 2: Total time taken

The total time is:

$\text{Total time} = t_1 + 2t = \frac{S}{6} + 2 \cdot \frac{S}{24}.$

Simplify:

$\text{Total time} = \frac{S}{6} + \frac{S}{12} = \frac{2S}{12} + \frac{S}{12} = \frac{3S}{12} = \frac{S}{4}.$

Step 3: Average speed

The average speed is given by:

$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}.$

Simplify:

$\text{Average speed} = \frac{2S}{\frac{S}{4}}.$

$\text{Average speed} = \frac{2S \cdot 4}{S} = 8 \, \text{m/s}.$

Final Answer: $8 \, \text{m/s}$.