Motion in One Dimension Practice Questions
HSC/SSC (Bangladesh) · HSC Physics · 138 free MCQs with instant results and detailed explanations.
138
Total
53
Easy
68
Medium
17
Hard
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Sample Questions from Motion in One Dimension
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Q1
Easy
An object moves in a straight line with a constant speed of 20 m/s. How far will it travel in 5 seconds?
A.
100 m
B.
80 m
C.
60 m
D.
120 m
Show Answer & Explanation
Correct Answer: A
Using the formula distance = speed × time, we calculate: distance = 20 m/s × 5 s = 100 m. Hence, the object travels 100 meters.
Q2
Easy
If a car accelerates from rest at a rate of 2 m/s², how fast will it be moving after 8 seconds?
A.
16 m/s
B.
24 m/s
C.
32 m/s
D.
8 m/s
Show Answer & Explanation
Correct Answer: A
Using the formula final velocity = initial velocity + (acceleration × time), we find: final velocity = 0 + (2 m/s² × 8 s) = 16 m/s.
Q3
Easy
If a train travels at a constant speed of 90 km/h, how far will it travel in 2 hours?
A.
90 km
B.
150 km
C.
180 km
D.
200 km
Show Answer & Explanation
Correct Answer: C
Distance is calculated using the formula: distance = speed × time. Here, distance = 90 km/h × 2 h = 180 km.
Q4
Medium
An object moves in a straight line with an initial velocity of 4 m/s and accelerates at -2 m/s². What is its velocity after 3 seconds?
A.
0 m/s
B.
2 m/s
C.
4 m/s
D.
10 m/s
Show Answer & Explanation
Correct Answer: B
Using the formula v = u + at, where u = 4 m/s, a = -2 m/s², and t = 3 s, we find v = 4 + (-2)*3 = 4 - 6 = -2 m/s, which indicates reverse direction.
Q5
Medium
An object moves in a straight line with a velocity of 12 m/s. If it accelerates at a rate of -2 m/s², what will be its velocity after 5 seconds?
A.
2 m/s
B.
12 m/s
C.
22 m/s
D.
7 m/s
Show Answer & Explanation
Correct Answer: A
To find the final velocity, we use the equation v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. Thus, v = 12 m/s + (-2 m/s² * 5 s) = 12 m/s - 10 m/s = 2 m/s.
Q6
Medium
A car starts from rest and accelerates uniformly at 2 m/s². How far does it travel in 5 seconds?
A.
25 meters
B.
10 meters
C.
50 meters
D.
15 meters
Show Answer & Explanation
Correct Answer: A
Using the formula for distance under uniform acceleration, s = ut + 0.5at², where u = 0, a = 2 m/s², and t = 5 s, we find s = 0 + 0.5 * 2 * (5)² = 25 meters.
Q7
Medium
A car accelerates uniformly from rest to a speed of 25 m/s in 10 seconds. What is the distance traveled by the car during this time?
A.
125 m
B.
250 m
C.
100 m
D.
150 m
Show Answer & Explanation
Correct Answer: A
The distance traveled can be calculated using the formula: Distance = initial velocity × time + 0.5 × acceleration × time². Here, initial velocity is 0, final velocity is 25 m/s, and time is 10 s. The acceleration is (25 m/s) / (10 s) = 2.5 m/s². Thus, distance = 0 + 0.5 × 2.5 × (10)² = 125 m.
Q8
Hard
A car accelerates uniformly from rest to a speed of 30 m/s in 10 seconds. What is the total distance covered by the car during this time?
A.
150 m
B.
120 m
C.
100 m
D.
200 m
Show Answer & Explanation
Correct Answer: A
Using the formula for distance covered under uniform acceleration, s = ut + (1/2)at². Here, u = 0, a = (v - u)/t = 3 m/s², and t = 10 s. Thus, s = 0 + (1/2)(3)(10)² = 150 m.
Q9
Hard
A car accelerates uniformly from rest to a speed of 30 m/s in 10 seconds. What is the total distance covered by the car during this time?
A.
150 m
B.
300 m
C.
450 m
D.
600 m
Show Answer & Explanation
Correct Answer: A
The distance covered under uniform acceleration can be calculated using the formula: distance = (initial velocity * time) + (1/2 * acceleration * time^2). With initial velocity = 0, we first find acceleration as 30 m/s divided by 10 s = 3 m/s². So, distance = (0 * 10) + (1/2 * 3 * 10^2) = 150 m.
Q10
Hard
An object is thrown vertically upward with an initial velocity of 40 m/s. What is the maximum height reached by the object? (Assume g = 10 m/s²)
A.
80 m
B.
160 m
C.
200 m
D.
320 m
Show Answer & Explanation
Correct Answer: B
The maximum height can be determined using the formula: height = (initial velocity²) / (2 * g). Here, height = (40 m/s)² / (2 * 10 m/s²) = 160 m. This shows the object rises until its velocity becomes zero due to gravity acting against it.
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Motion in One Dimension — HSC/SSC (Bangladesh) HSC Physics Practice Questions Online
This page contains 138 practice MCQs for the chapter Motion in One Dimension in HSC/SSC (Bangladesh) HSC Physics. The questions are organized by difficulty — 53 easy, 68 medium, 17 hard — so you can choose the right level for your preparation.
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