(a) The area under the acceleration vs time graph gives the effect of acceleration through time, which is a speed change Δv.
Δv = final speed - initial speed = v - 0 = 2t
The speed at any time, t, is: v = 2t
Which gives a speed vs time graph as a straight line from (0,0) to (t,2t).

The speed at 5s is given by v = 2 × 5 = 10 m.s^-1.




(b) The change in height will be the triangular area under the speed vs time graph.
Δx = final position - initial position = x - 0 = ½×t×2t = t².
The position at any time, t, is: x = t².
The position vs time graph is a curve from (0,0) to (t,t²).

The height at 10s will be x = 10² = 100 m.

(a) The acceleration vs time graph is a horizontal line at +5m.s^-2.
The area at any time, t, under the acceleration time graph is Δv = 5t, i.e. v - 4 = 5t
The speed at any time, t, is v = 4 + 5t.
At t = 0s, the speed is 4 m.s^-1.
At t = 2s, the speed is 14 m.s^-1.

(b) Which gives a speed vs time graph as a straight line from (0,4) to (2,14).
The area under the speed vs time graph gives the distance travelled.
The area is a rectangle (4 × 2) plus a triangle (½×2×10)
= 8 + 10 = 18m.
The rectangle gives the distance that would have been covered if moving at the constant initial speed.
The triangle gives the extra distance covered due to the acceleration.

(a) Which gives a speed vs time graph as a straight line from (0,2) to (30,8).
The average acceleration is the change in speed divided by the time during which the speed changed.
The average acceleration is (8 - 2)/30 = 0.2m.s^-2.

(b) The distance travelled is the area under the speed vs time graph.
The distance travelled is the area of the rectangle (30 × 2) plus the area of the triangle (½×30×6) = 60 + 90 = 150m

(a) The change in speed is the area under the acceleration vs time graph, i.e. Δv = v = -1.6t
Which gives a speed vs time graph as a straight line from (0,0) to (t, -1.6t).
The distance travelled is the area under the speed vs time graph, i.e. Δh = h - 5 = ½×(-1.6t)×t.
h = 5 - 0.8×t²
The height graph is then a curve down from 5m to the ground.
(a) To find the time it takes to hit the ground, ask the question: what is t when h=0?
The time will not be any time, t, but a particular time t₀ (a particular time is indicated by a subscript).
0 = 5 - 0.8×t₀²
t₀² = 5/0.8 = 6.25, so t₀ = ±2.5s
A negative time occurred before the start of the motion and so is discarded, hence t₀ = 2.5s.

(b) The speed just before it hits the surface is the speed, v₀, at time t₀
v₀ = -1.6×2.5 = -4m.s^-1. The minus sign give the direction of the speed as toweards the ground.

(a) The change in speed is the area under the acceleration vs time graph, i.e. Δv = v = -26t
Which gives a speed vs time graph as a straight line from (0,0) to (t, -26t).
The final conditions give a particular speed v₀ = -36.7 km/h at the particular time t₀.
In SI units this is -36.7×1000/3600 m.s^-1 = -10.2m.s^-1.
Substituting in the speed vs time equation gives
-10.2 = -26·t₀   or  t₀ = 0.4s.

(b) The change in the height is the area under the speed vs time graph, i.e. Δh = h - h₀ = ½·(-26t)·t
h = h₀ -13·t²
Substituting h = 0m at t₀ = 0.4s gives:
h₀ = 13×(0.4)² = 2.08m.

The change in speed is the area under the acceleration vs time graph, i.e. Δv = v = at
Which gives a speed vs time graph as a straight line from (0,0) to (t, at).
The final conditions give a particular speed v₀ = 360 km/h at the particular time t₀ when it is 1800m away.
In SI units this is 360×1000/3600 m.s^-1 = 100m.s^-1.
Substituting into the speed vs time equation gives
100 = a·t₀
The change in position is the area under the speed vs time graph, i.e. Δx = x = ½·(at)·t
x = ½at²
Substituting x = 1800m at t₀ = 100/a gives:
1800 = ½a·(100/a)² = 100²/2a
Rearranging: a = 100²/3600 = 2.8 m.s^-2
Which gives t₀ = 100/2.8 = 36s

The change in speed is the area under the acceleration vs time graph, i.e. Δv = -9.8t or v = u - 9.8t
Which gives a speed vs time graph as a straight line from (0,u) to (t₀,0) and u = 9.8t₀
The change in position is the area under the speed vs time graph, i.e. Δx = ½·ut₀

Substituting Δx = 19.6m at t₀ = u/9.8 gives:
19.6 = ½·u·(u/9.8)
Re-arranging: u² = (19.6)², i.e. u = ±19.6m.s^-1
A negative speed does not fit the conditions and so is discarded,
hence u = 19.6m.s^-1.
Substituting: t₀ = u/9.8 = 19.6/9.8 = 2s

The change in speed is the area under the acceleration vs time graph, i.e. Δv = at or v = u + at
Which gives a speed vs time graph as a straight line from (0,u) to (t₀,v).
The change in position is the area under the speed vs time graph, i.e. Δx = ut₀ + ½·(v - u)t₀
Re-arranging: Δx = ½·(v + u)t₀
Substituting Δx = 60m and v = 16 m.s^-1 at 6s gives:
60 = ½·(16 + u)×6
Re-arranging: u +16 = 20, i.e. u = 4m.s^-1
From the speed vs time equation: v = u + at
Substituting: 16 = 4 + a×6, i.e. a = 2m.s^-2

With this information, new graphs can be drawn for the time before the two points were passed.

The rectangular area from -t₀ to the origin under the acceleration vs time graph gives the change in speed from 0m.s^-1 to 4m.s^-1. Using this: Δv = -2t₀ = 4, i.e. t₀ = -2s

The triangular area from -t₀ to the origin under the speed vs time graph gives the change in position.
Δx = ½×4×2 = 4m.

The change in the bus's speed is an area under the acceleration vs time graph, i.e. Δv_b = v_b = 3t
Which gives the bus's speed vs time graph as a straight line from (0,0) to (t,3t).
The change in the student's speed is zero (constant speed, zero acceleration).
Which gives the student's speed vs time graph as a horizontal line, i.e. v_s = 6.
The change in the bus's position is an area under the speed vs time graph, i.e. Δx_b = ½·(3t)t, or x_b = 12 + 1.5t²
The change in the student's position is an area under the speed vs time graph, i.e. Δx_s = x_s = 6t
For the student to catch the bus, there must be a time when x_b = x_s
Set 12 + 1.5t² = 6t
Re-arranging: t² - 4t + 8 = 0
This is a quadratic equation and the discriminant will tell whether there are 2 real solutions, 1 real solution, or no real solutions.
The discriminant is 4² - 4×1×8 = -16
The negative disciminant shows that situation 2 applies, there is no time when the bus and the student are at the same place.

The cyclist finishes the race at time t₂
The runner finishes the race at time t₃
The graphs are for illustration only and t₂ could be greater than t₃

The change in the cyclist's speed is an area under the acceleration vs time graph, i.e. Δv_c = v_c = 5t, 0 < t ≤ t₁
When t = t₁ the cyclist's speed reaches v_c = 20 m.s^-1.
Substituting this gives 20 = 5t₁ ∴ t₁ = 4s
The cyclist's speed is constant at 20 m.s^-1, 4s ≤ t ≤ t₂
The change in the runner's speed is zero (constant speed, zero acceleration).
Which gives the runners's speed as v_r = 10, 0 < t ≤ t₃

The change in the cyclist's position is an area under the speed vs time graph, i.e. Δx_c = x_c = ½·20×4 + 20×(t₂ - 4)
Re-arranging: x_c = 20t₂ - 40
At t = t₂ the cyclist has finished the race and x_c = 50m
Substituting this gives 50 = 20t₂ - 40, i.e. t₂ = 4.5s

The change in the runner's position is an area under the speed vs time graph, i.e. Δx_r = x_r = 10t₃
At t = t₃ the runner has finished the race and x_r = 50m
Substituting this gives 50 = 10t₃, i.e. t₃ = 5s
The runner finishes 0.5s behind the cyclist.

When x is 39.6m, u = 64.8 km/h = 64.8×1000/3600 = 18m.s^-1
When x is 64.8m, u = 86.4 km/h = 86.4×1000/3600 = 24m.s^-1
The reaction time is t₁, and in each case, up to this time the car travels at constant speed. The time when stopped is t₂.
The change in speed is the area under the acceleration vs time graph, i.e. Δv = v - u = - a(t₂ - t₁), t₁ ≤ t ≤ t₂
When stopped, v = 0 ∴ (t₂ - t₁) = u/a

The change in position is the area under the speed vs time graph, i.e. Δx = x = ut₁ + ½·u(t₂ - t₁)

Substituting (t₂ - t₁) = u/a from above, gives:

When x = 39.6m, u = 18m.s^-1
When x = 64.8m, u = 24m.s^-1 - substituting:

dividing by 18 and 24:

subtracting: 0.5 = 3/a,   i.e. a = 6m.s^-2
substituting: 2.7 = t₁ + 12/6,   i.e. t₁ = 0.7s