Class 11 Physics Chapter 2: Motion in a Straight Line worksheet with answer key
A 20-mark worksheet on Motion in a Straight Line for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (c) 40 m
Displacement is the area under the v-t line, a trapezium: ½ (4 + 12) × 5 = 40 m. Using only the final velocity (12 × 5) gives 60 m and only the initial velocity gives 20 m.
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Answer: (d) −5 m s⁻²
ā = (v₂ − v₁)/(t₂ − t₁) = (−4 − 6)/2 = −5 m s⁻², that is, 5 m s⁻² towards the west. Taking −1 m s⁻² ignores the reversal of direction.
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Answer: (a) 100 m
Let the westward run be w. Path length = 300 + w; magnitude of displacement = 300 − w. Both are divided by the same time, so 300 + w = 2(300 − w), 3w = 300 and w = 100 m.
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Answer: (b) the size of the bus is much smaller than the distance it moves
A body may be treated as a point object when its size is much smaller than the distance it moves in the time considered; a bus about 10 m long covering 60 km qualifies. Speed, mass and the shape of the road do not decide this.
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Answer: (b) negative
The slope of the x-t graph is the velocity. A curve bending downward has a decreasing slope, so the velocity decreases and the acceleration is negative; a straight line would mean zero acceleration.
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(i) Air resistance is negligible, so the only acceleration is that due to gravity (free fall). (ii) The height of fall is small compared with the earth's radius, so g can be taken as constant, 9.8 m s⁻².
- air resistance negligible (1)
- height small compared with the earth's radius, so g constant (1)
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For constant acceleration, x = ½(v₀ + v)t: 60 = ½(v₀ + 20) × 4, so v₀ + 20 = 30 and v₀ = 10 m s⁻¹. From v = v₀ + at, a = (20 − 10)/4 = 2.5 m s⁻².
- v₀ = 10 m s⁻¹ from the average-velocity relation (1)
- a = 2.5 m s⁻² (1)
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a = dv/dt = 6t, so v = ∫₀ᵗ 6t dt = 3t²; at t = 2 s, v = 12 m s⁻¹. v = dx/dt = 3t², so x = ∫₀ᵗ 3t² dt = t³; at t = 2 s, x = 8 m. The equations for constant acceleration cannot be used, since a changes with time.
- v = 3t², 12 m s⁻¹ (1)
- x = t³, 8 m (1)
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(a) Both balls have the same acceleration g, so the velocity of B relative to A stays 20 m s⁻¹ upward. The 40 m gap closes in t = 40/20 = 2 s. (b) A has fallen ½gt² = 5 × 4 = 20 m, so they meet 40 − 20 = 20 m above the ground. (c) vB = 20 − 10 × 2 = 0: B is at its highest point when they meet.
- relative velocity constant, t = 2 s (1)
- meeting height 20 m (1)
- vB = 0, B at its highest point (1)
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The v-t graph is a straight line from (0, v₀) to (t, v). (a) The area under it is a rectangle of height v₀ and base t plus a triangle of base t and height (v − v₀): x = v₀t + ½(v − v₀)t. Since v − v₀ = at, x = v₀t + ½at². (b) The area is also a trapezium: x = ½(v + v₀)t. With t = (v − v₀)/a, x = (v + v₀)(v − v₀)/2a = (v² − v₀²)/2a, so v² = v₀² + 2ax.
- area under the v-t graph equals displacement; area split into rectangle and triangle (1)
- substituting v − v₀ = at to get x = v₀t + ½at² (1)
- trapezium area with t = (v − v₀)/a gives v² = v₀² + 2ax (1)
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(a) From rest, x = ½at²: 4 = ½a(1)², a = 8 cm s⁻². (b) sₙ = (a/2)(2n − 1) = 4(2n − 1); for n = 4, s₄ = 4 × 7 = 28 cm (the distances in successive seconds go as 1 : 3 : 5 : 7). (c) 4(2n − 1) = 44 gives 2n − 1 = 11, n = 6: the sixth second.
- a = 8 cm s⁻² (1)
- 28 cm in the fourth second (1)
- sixth second (1)
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