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Difficulty: MediumKinematics and Linear Motion

A car starts from rest and accelerates uniformly to a velocity of 16 m/s16\text{ m/s} in 4 s4\text{ s}. It continues at this constant velocity for 8 s8\text{ s}, and then decelerates uniformly to rest in another 4 s4\text{ s}. What is the average speed of the car for the entire journey?

  1. 12 m/s12\text{ m/s}Answer
  2. B
    16 m/s16\text{ m/s}
  3. C
    8 m/s8\text{ m/s}
  4. D
    24 m/s24\text{ m/s}

Answer

12 m/s12\text{ m/s}
The average speed of an object undergoing multi-stage motion is defined as the total distance traveled divided by the total time taken. The total distance covered is 32 m32\text{ m} (during acceleration) +128 m+ 128\text{ m} (during constant velocity) +32 m+ 32\text{ m} (during deceleration) =192 m= 192\text{ m}. Dividing 192 m192\text{ m} by the total time of 16 s16\text{ s} yields 12 m/s12\text{ m/s}.

Step-by-Step Solution

1
Calculate the distance covered in each stage of motion.
Acceleration stage (s1s_1): 0+162×4=32 m\frac{0 + 16}{2} \times 4 = 32\text{ m}. Constant velocity stage (s2s_2): 16×8=128 m16 \times 8 = 128\text{ m}. Deceleration stage (s3s_3): 16+02×4=32 m\frac{16 + 0}{2} \times 4 = 32\text{ m}.
The total distance is the sum of distances traveled during acceleration, constant speed, and deceleration.
2
Find total distance traveled and total time taken.
Total distance (SS) = 32 m+128 m+32 m=192 m32\text{ m} + 128\text{ m} + 32\text{ m} = 192\text{ m}. Total time (TT) = 4 s+8 s+4 s=16 s4\text{ s} + 8\text{ s} + 4\text{ s} = 16\text{ s}.
Average speed requires total distance divided by total elapsed time.
3
Compute the average speed.
Average speed = ST=192 m16 s=12 m/s\frac{S}{T} = \frac{192\text{ m}}{16\text{ s}} = 12\text{ m/s}.
Dividing total distance by total time gives the average speed over the entire motion.

Key Concept

Average Speed in Multi-Stage Motion
Estimated Time:1m 30s
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Topics

Kinematics and Linear Motion

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