Motion in One Dimension Quiz - ICSE Class 9 Physics
Free practice quiz: Motion in One Dimension (ICSE Class 9 Physics). 67 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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A car accelerates uniformly from rest for 5 seconds. If its final velocity is 20 m/s, what is its acceleration?
- 2 m/s²
- 4 m/s²
- 5 m/s²
- 10 m/s²
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Answer: B. 4 m/s²
Using the first equation of motion, v = u + at. Given u = 0 m/s (rest), v = 20 m/s, and t = 5 s. Therefore, 20 = 0 + a * 5, which gives a = 20/5 = 4 m/s².
An object starts from rest and accelerates uniformly at 3 m/s² for 10 seconds. What is the distance covered during this time?
- 150 m
- 300 m
- 100 m
- 200 m
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Answer: A. 150 m
Using the equation s = ut + (1/2)at². Given u = 0 m/s (rest), a = 3 m/s², t = 10 s. So, s = (0)(10) + (1/2)(3)(10)². This simplifies to s = (1/2)(3)(100) = 150 m.
A car starts from rest and accelerates uniformly for 10 seconds, reaching a speed of 20 m/s. It then travels at a constant speed for 15 seconds before decelerating uniformly to rest in 5 seconds. What is the total distance traveled?
- 350 m
- 450 m
- 500 m
- 550 m
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Answer: B. 450 m
The motion can be divided into three phases: Phase 1 (Acceleration): Initial velocity (u) = 0 m/s, final velocity (v) = 20 m/s, time (t) = 10 s. The acceleration (a) = (v - u) / t = (20 - 0) / 10 = 2 m/s². The distance covered (s1) = ut + (1/2)at² = 0(10) + (1/2)(2)(10)² = 100 m. Phase 2 (Constant Speed): The car travels at a constant speed of 20 m/s for 15 seconds. The distance covered (s2) = speed × time = 20 m/s × 15 s = 300 m. Phase 3 (Deceleration): Initial velocity (u) = 20 m/s, final velocity (v) = 0 m/s, time (t) = 5 s. The acceleration (a) = (v - u) / t = (0 - 20) / 5 = -4 m/s². The distance covered (s3) = ut + (1/2)at² = 20(5) + (1/2)(-4)(5)² = 100 - 50 = 50 m. Total distance traveled = s1 + s2 + s3 = 100 m + 300 m + 50 m = 450 m.
An object starts from rest and undergoes uniform acceleration. If it covers 100 meters in 10 seconds, what is its acceleration?
- 1 m/s²
- 2 m/s²
- 3 m/s²
- 4 m/s²
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Answer: B. 2 m/s²
Using the equation of motion: s = ut + (1/2)at². Given distance (s) = 100 m, initial velocity (u) = 0 m/s (starts from rest), and time (t) = 10 s. Substituting these values: 100 = (0)(10) + (1/2)a(10)². This simplifies to 100 = (1/2)a(100), so 100 = 50a. Therefore, a = 100 / 50 = 2 m/s².
A person walks 100 m towards the east in 60 seconds. What is their average velocity?
- 100/60 m/s east
- 60/100 m/s east
- 100/60 m/s west
- 60/100 m/s west
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Answer: A. 100/60 m/s east
Average velocity is defined as total displacement divided by total time. The displacement is 100 m towards the east, and the total time is 60 seconds. Therefore, the average velocity is (100 meters East) / (60 seconds) = 100/60 m/s East.