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Show that an object dropped from rest near the Earth’s surface falls 19.6 m in 2.0 s. Take g = 9.81 m s⁻².
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Using s = ut + ½at² with u = 0: $$s=ut+\frac{1}{2}a{t}^{2}$$ $$s=\frac{1}{2}\times 9.81\times {2.0}^{2}$$ $$s=19.62\text{ m}\approx 19.6\text{ m}$$ as required.
A train brakes from 30 m s⁻¹ to rest with a uniform deceleration of magnitude 1.2 m s⁻². Calculate the braking distance.
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Taking the initial direction as positive, a = –1.2 m s⁻². $${v}^{2}={u}^{2}+2as$$ $$s=\frac{{v}^{2}-{u}^{2}}{2a}$$ $$s=\frac{0-{30}^{2}}{2\times (-1.2)}$$ $$s=375\text{ m}$$
A cyclist accelerates uniformly from rest. State what happens to the distance travelled in a fixed time if the acceleration is doubled.
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from s = ½at², s is directly proportional to a for fixed t; doubling a doubles the distance travelled
A ball is thrown vertically upwards at 14 m s⁻¹ from a point 1.8 m above the ground. Take g = 9.81 m s⁻² and ignore air resistance. Determine the speed at which it strikes the ground.
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Taking upwards as positive: u = +14 m s⁻¹, a = –9.81 m s⁻², s = –1.8 m (ground is below launch point). $${v}^{2}={u}^{2}+2as$$ $${v}^{2}={14}^{2}+2\times (-9.81)\times (-1.8)$$ $${v}^{2}=196+35.3=231.3$$ $$v=15.2{\text{ m s}}^{-1}$$ Speed on impact ≈ 15 m s⁻¹.
State what a horizontal line on a velocity-time graph represents.
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constant velocity / zero acceleration
Describe the motion shown by a curved displacement-time graph that becomes steeper with time.
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the gradient is increasing; therefore the velocity is increasing / the object is accelerating in the positive direction
A car travels at constant velocity, then brakes uniformly to rest, then remains stationary. Sketch the corresponding velocity-time graph and acceleration-time graph for the whole journey.
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velocity-time: horizontal line at v₀, then straight line with negative gradient down to zero, then horizontal line on the time axis; acceleration-time: horizontal line at zero, then horizontal line at constant negative value, then horizontal line at zero; transitions aligned vertically between the two graphs
State what is given by the area under an acceleration-time graph.
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the change in velocity
A car decelerates from 30 m s⁻¹ at 5.0 m s⁻². Deduce whether the car can stop within 100 m.
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Taking the initial direction as positive, a = –5.0 m s⁻². $${v}^{2}={u}^{2}+2as$$ $$s=\frac{{v}^{2}-{u}^{2}}{2a}$$ $$s=\frac{0-{30}^{2}}{2\times (-5.0)}=90\text{ m}$$ 90 m < 100 m, so the car does stop within 100 m.
Explain how the instantaneous velocity is determined from a curved displacement-time graph at a particular time.
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a tangent is taken to the curve at the chosen time; the gradient of that tangent is calculated using Δs/Δt over a large triangle; this gradient equals the instantaneous velocity at that point
A driver travelling at 25 m s⁻¹ sees an obstacle and applies the brakes. The reaction time is 0.70 s and the braking deceleration is 6.5 m s⁻². Discuss, with calculation, whether the car stops within 60 m of the obstacle. In your answer you should: • calculate the distance travelled during the reaction time • calculate the braking distance using uniformly accelerated motion • evaluate whether the total stopping distance exceeds 60 m.
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During reaction time, velocity is constant: $${s}_{1}=vt=25\times 0.70$$ $${s}_{1}=17.5\text{ m}$$ During braking, taking initial direction as positive: u = 25 m s⁻¹, v = 0, a = –6.5 m s⁻²: $${v}^{2}={u}^{2}+2as$$ $${s}_{2}=\frac{{v}^{2}-{u}^{2}}{2a}$$ $${s}_{2}=\frac{0-{25}^{2}}{2\times (-6.5)}=48.1\text{ m}$$ Total stopping distance: $${s}_{\text{total}}=17.5+48.1=65.6\text{ m}$$ 65.6 m > 60 m, so the car does not stop within 60 m of the obstacle.