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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.

Taking the initial direction as positive, a = –1.2 m s⁻². $${v}^{2}={u}^{2}+2as$$ $$s=\frac{{v}^{2}-{u}^{2}}{2a}$$ $$s=\frac{0-{30}^{2}}{2\times (-1.2)}$$ $$s=375\text{ m}$$

from s = ½at², s is directly proportional to a for fixed t; doubling a doubles the distance travelled

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.

constant velocity / zero acceleration

the gradient is increasing; therefore the velocity is increasing / the object is accelerating in the positive direction

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

the change in velocity

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.

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

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.

a scalar has only magnitude / a vector has both magnitude and direction; example of scalar (e.g. mass / energy / time) and example of vector (e.g. force / velocity / displacement)

weight; momentum; acceleration (all three required for both marks)

Finding the horizontal component $${F}_{x}=120cos40^{\circ}=91.9\text{ N}$$ Finding the vertical component $${F}_{y}=120sin40^{\circ}=77.1\text{ N}$$

The horizontal (6.0 m s⁻¹) and vertical (8.0 m s⁻¹) components are perpendicular, so combine them with Pythagoras for the magnitude and use the tangent ratio for the direction.

Magnitude: v = √(vₓ² + v_y²) = √(6.0² + 8.0²) = √(36 + 64) = √100 = 10 m s⁻¹

Direction (angle above the horizontal): θ = tan⁻¹($v_{\mathrm{y}}$ / vₓ) = tan⁻¹(8.0 / 6.0) = tan⁻¹(1.333) = 53° above the horizontal (53.1°; equivalently 37° from the vertical)

Result: the velocity has a magnitude of 10 m s⁻¹ directed at 53° above the horizontal.

Because the two forces are perpendicular, they form the two legs of a right-angled triangle, so the magnitude of the resultant is found with Pythagoras' theorem:

R = √(9.0² + 12.0²) R = √(81 + 144) R = √225 R = 15 N

The magnitude of the resultant force is 15 N. (This is a 3-4-5 right triangle scaled ×3.) If a direction is required, it acts at θ = tan⁻¹(9.0/12.0) ≈ 37° from the 12.0 N force (equivalently tan⁻¹(12.0/9.0) ≈ 53° from the 9.0 N force).

$$R=\sqrt{{24}^{2}+{7.0}^{2}}$$ $$R=\sqrt{576+49}=\sqrt{625}=25{\text{ m s}}^{-1}$$ as required.

there is no horizontal force acting on the projectile (only gravity acts vertically); therefore the horizontal acceleration is zero, so horizontal velocity stays constant

Taking downwards as positive, for the vertical motion: $$s=ut+1/2a{t}^{2} (u=0)$$ $$45=1/2\times 9.81\times {t}^{2}$$ $${t}^{2}=\frac{90}{9.81}=9.174$$ $$t=3.03\text{ s}$$

$$R={u}_{x}\times t=15\times 3.03$$ $$R=45.4\text{ m}\approx 45\text{ m}$$ as required.