Motion in One Dimension Exe-2C MCQs Class-9 Concise Physics ICSE Selina Solutions. In this article you would learn about equation of motions. Visit official Website CISCE for detail information about ICSE Board Class-9.
Motion in One Dimension Exe-2C MCQs Class-9 Concise Physics ICSE Selina Solutions Ch-2
| Board | ICSE |
| Class | 9 |
| Subject | Physics |
| Writer / Publication | Concise selina Publishers |
| Chapter-2 | Motion in One Dimension |
| Exe – 2 C | Equations of Motion |
| Topics | Solution of Exe-2(C) MCQs Type |
| Academic Session | 2025-2026 |
Equations of Motion
Motion in One Dimension Exe-2C MCQs Class-9 Concise Physics ICSE Selina Solutions Ch-2
Page 59 (Choose the correct answer from the options given below).
Que-1: The correct equation of motion is :
(a) v = u + at
(b) S = ut +1/2 at²
(c) v² = u² + 2aS
(d) All of these.
Ans: (d) All of these
Hint- For motion of a body moving with a uniform acceleration, the following three equations give the relationship between initial velocity (u), final velocity (v), acceleration (a), time of journey (t) and distance travelled (S) :
v = u + at
S = ut +1/2 at²
v² = u² + 2aS
Que-2: When a body starts from rest, the equation of motion takes the form:
(a) v = u
(b) v = at
(c) v = 1/2 at²
S = ut + 1/2 at²
Ans: (b) v = at
Hint- When a body starts from rest, initial velocity is zero (u = 0), then v = at.
Que-3: If a body is moving with a uniform retardation, then its acceleration will be:
(a) Positive
(b) Negative
(c) Zero
(d) Cannot say
Ans: (b) Negative
Hint- When a body is moving with uniform retardation, then a will be negative.
Que-4: A car acquires a velocity of 54 ms-1 in 20 s starting from rest, then its acceleration is:
(a) 5.4 ms-2
(b) 2.7 ms-2
(c) 7.2 ms-2
(d) 2.0 ms-2
Ans: (b) 2.7 ms-2
Hint- Given,
u = 0
v = 54 ms-1
t = 20 s
a = ?
Using, v = u + at or
a = (v−u)/t
a = (54−0)/20
= 2.7 ms-2.
Hence, acceleration = 2.7 ms-2
Que-5: A particle starts to move in a straight line from a point with a velocity 10 ms-1 and acceleration -2.0 ms-2. Its position at t = 5 s will be :
(a) 5 m
(b) 10 m
(c) 20 m
(d) 25 m
Ans; (d) 25 m
Hint- Given,
u = 10 ms-1
a = -2.0 ms-2
t = 5 s
Using, S = ut + 1/2 at2
S = (10 x 5) + 1/2 x (-2) x 52 = [50 – 25] = 25 m
Hence, its position at t = 5 s will be 25 m
Que-6: A body initially at rest, starts moving with a constant acceleration of 0.5 ms-2 and travels a distance 25 m, then its final velocity is:
(a) 5 ms-1
(b) 20 ms-1
(c) 15 ms-1
(d) -15 ms-1
Ans: (a) 5 ms-1
Hint- Given,
u = 0
a = 0.5 ms-2
S = 25 m
v = ?
Using, v2 = u2 + 2aS
v2 = 0 + [2 x 0.5 x 25] = 25
Hence, v = 5 ms-1
Que-7: A body starts from rest with a uniform acceleration of 8 ms-2, then the distance covered by the body in 2 s is:
(a) 4 m
(b) 16 m
(c) 20 m
(d) 32 m
Ans: (b) 16 m
Hint- Given,
u = 0
a = 8 ms-2
t = 2 s
S = ?
Using, S = ut + 1/2at2
S = (0 x 2) + 1/2 x 8 x 22 = 16 m
Hence, the distance covered by the body is 16 m
Page 60
Que-8: A car starting from rest accelerates uniformly to acquire a speed 20 km h-1 in 30 min. The distance travelled by car in this time interval will be —
(a) 600 km
(b) 5 km
(c) 6 km
(d) 10 km
Ans: (b) 5 km
Hint- As we know,
v = u + at
Given,
u = 0
v = 20 km h-1
t = 30 min = 1/2 h
Substituting the values in the formula above, we get,
20 = 0+(a×1/2)
⇒ a = 40 km h-2
Now,
S = ut + 1/2 at²
Substituting the values in the formula above we get,
S=(0×0.5)+[(1/2)×40×(1/2)²]
S=0+{(1/2)×40×(1/4)}
⇒S=5 km
Hence, distance = 5 km
— : End of Motion in One Dimension Exe-2C MCQs Class-9 Concise Physics ICSE Selina Solutions. :–
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