The techniques of finding slopes (how quickly something changes in time or space) and
finding areas (the effect of something through time or space) will now be applied to position, speed and acceleration.
Example problems. Fully worked answers are provided.
Tutorial questions
1. A hot air balloon rises vertically from rest on the ground with a constant acceleration of 2 m.s^{2}. Find
(a) the speed reached after 5 s.
(b) the height reached after 10 s.

(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 = t^{2}.
The position at any time, t, is: x = t^{2}.
The position vs time graph is a curve from (0,0) to (t,t^{2}).
The height at 10s will be x = 10^{2} = 100 m. 
2. A submarine cruising at 4 m.s^{1} accelerates at 5 m.s^{2} for 2 second. Find
(a) the speed reached after the 2 s. (ans 14ms^{1})
(b) the distance travelled while accelerating. (ans 18m)

(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. 
3. A bicycle rider uniformly accelerates to 28.8 km.hr^{1} from 7.2 km.hr^{1} in half a
minute. Find
(a) the average acceleration during this time. (ans 0.2m.s^{2})
(b) the distance travelled during this time. (ans 150m)

(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 
4. When an object is released on the moon it will accelerate towards the surface at 1.6m.s^{2}.
A hammer is released from rest at 5m above the surface when the time is zero second. Find
(a) the time it takes to hit the surface. (ans 2.5s)
(b) the speed just before it hits the surface. (ans 4m.s^{1})

(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×t^{2}
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 t_{0} (a particular time is indicated by a subscript).
0 = 5  0.8×t_{0}^{2}
t_{0}^{2} = 5/0.8 = 6.25, so t_{0} = ±2.5s
A negative time occurred before the start of the motion and so is discarded, hence t_{0} = 2.5s.
(b) The speed just before it hits the surface is the speed, v_{0}, at time t_{0}
v_{0} = 1.6×2.5 = 4m.s^{1}. The minus sign give the direction of the speed as toweards the ground.

5. The gravitational acceleration on Jupiter is estimated at 26 m.s^{2}.
A lead ball falls from rest and acquires a speed of 36.7 km.hr^{1} just before it hits the surface of Jupiter. Find
(a) the time taken in the fall. (ans 0.39s)
(b) the height from which the lead ball fell to give it this speed. (ans 2.08m)

(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 v_{0} = 36.7 km/h at the particular time t_{0}.
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·t_{0} or t_{0} = 0.4s.
(b) The change in the height is the area under the speed vs time graph, i.e.
Δh = h  h_{0} = ½·(26t)·t
h = h_{0} 13·t^{2}
Substituting h = 0m at t_{0} = 0.4s gives:
h_{0} = 13×(0.4)^{2} = 2.08m.

6. A Jumbo jet is taken to have a constant acceleration from rest when it takes off.
It has 1.8 km of runway in which to take off and needs to reach a speed of 360 km.hr^{1}. Find
(a) the minimum acceleration needed to take off. (ans 2.8m.s^{1})
(b) the time taken during this minimum acceleration. (ans 36s)

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 v_{0} = 360 km/h at the particular time t_{0} 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·t_{0}
The change in position is the area under the speed vs time graph, i.e.
Δx = x = ½·(at)·t
x = ½at^{2}
Substituting x = 1800m at t_{0} = 100/a gives:
1800 = ½a·(100/a)^{2} = 100^{2}/2a
Rearranging: a = 100^{2}/3600 = 2.8 m.s^{2}
Which gives t_{0} = 100/2.8 = 36s

7. In an 80 km/hr speed restriction zone, a motorist makes an emergency stop with the brakes immediately locking
and the wheels sliding. The length of the skid marks on the road were 19.6 m.
A police officer assumes that the maximum deceleration of the car would not exceed the acceleration of a freely falling body
(i.e. 9.8 m.s^{2}) and "books" the motorist for speeding.
Using the police officer's assumption, find
(a) if the motorist was exceeding the speed limit. (ans 2s)
(b) how long it took for the car to come to rest. (ans no, 19.6m.s^{1} <22.2m.s^{1},
but the assumption was not good because you often decelerate in a car at less than "g".)

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 (t_{0},0) and u = 9.8t_{0}
The change in position is the area under the speed vs time graph, i.e.
Δx = ½·ut_{0}
Substituting Δx = 19.6m at t_{0} = u/9.8 gives:
19.6 = ½·u·(u/9.8)
Rearranging: u^{2} = (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: t_{0} = u/9.8 = 19.6/9.8 = 2s

8. A train moving with constant acceleration travels between two points 60 m apart in 6 s.
Its speed as it passes the second point is 16 m.s^{1}. Find
(a) the speed of the train at the first point. (ans 4m.s^{1})
(b) the acceleration during this time. (ans 2m.s^{1})
(c) the distance before the first point where it was stationary. (ans 4m)

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 (t_{0},v).
The change in position is the area under the speed vs time graph, i.e.
Δx = ut_{0} + ½·(v  u)t_{0}
Rearranging: Δx = ½·(v + u)t_{0}
Substituting Δx = 60m and v = 16 m.s^{1} at 6s gives:
60 = ½·(16 + u)×6
Rearranging: 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 t_{0} 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 = 2t_{0} = 4, i.e. t_{0} = 2s
The triangular area from t_{0} to the origin under the speed vs time graph gives
the change in position.
Δx = ½×4×2 = 4m.

9. A student runs toward a stationary bus at a steady speed of 6 m.s^{1}.
When the student is 12 m from the bus, it begins moving away from the student with a
constant acceleration of 3 m.s^{2}.
Find out if the student can catch the bus without altering speed.

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 + 1.5t^{2}
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 + 1.5t^{2} = 6t
Rearranging: t^{2}  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 4^{2}  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.

10. A runner and a cyclist have a race over a 50 m track.
The runner keeps up a steady pace of 10 m.s^{1} for the entire distance.
The cyclist accelerates uniformly from rest at 5 m.s^{2} to a maximum speed of 20 m.s^{1}
that is then maintained for the remainder of the race.
Find out who crossed the finish line first.
The cyclist finishes the race at time t_{2}
The runner finishes the race at time t_{3}
The graphs are for illustration only and t_{2} could be greater than t_{3}

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 ≤ t_{1}
When t = t_{1} the cyclist's speed reaches v_{c} = 20 m.s^{1}.
Substituting this gives 20 = 5t_{1} ∴ t_{1} = 4s
The cyclist's speed is constant at 20 m.s^{1}, 4s ≤ t ≤ t_{2}
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 ≤ t_{3}
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×(t_{2}  4)
Rearranging: x_{c} = 20t_{2}  40
At t = t_{2} the cyclist has finished the race and x_{c} = 50m
Substituting this gives 50 = 20t_{2}  40, i.e. t_{2} = 4.5s
The change in the runner's position is an area under the speed vs time graph, i.e.
Δx_{r} = x_{r} = 10t_{3}
At t = t_{3} the runner has finished the race and x_{r} = 50m
Substituting this gives 50 = 10t_{3}, i.e. t_{3} = 5s
The runner finishes 0.5s behind the cyclist.

11. A car with good brakes moving at 64.8 km/hr stops in 39.6 m.
The corresponding distance when moving at 86.4 km/hr is 64.8 m.
Assume that the reaction time of the driver is the same for both situations.
Assume that the deceleration of the car is the same for both situations. Find
(a) the deceleration of the car. (ans 6m.s^{2})
(b) the reaction time of the driver. (ans 0.7s)

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 t_{1}, and in each case, up to this time the car travels at constant speed.
The time when stopped is t_{2}.
The change in speed is the area under the acceleration vs time graph,
i.e. Δv = v  u =  a(t_{2}  t_{1}), t_{1} ≤ t ≤ t_{2}
When stopped, v = 0 ∴ (t_{2}  t_{1}) = u/a
The change in position is the area under the speed vs time graph, i.e.
Δx = x = ut_{1} + ½·u(t_{2}  t_{1})
Substituting (t_{2}  t_{1}) = u/a from above, gives:
When x = 39.6m, u = 18m.s^{1}
When x = 64.8m, u = 24m.s^{1}  substituting:
39.6 = 18t_{1} + 18^{2}/2a
64.8 = 24t_{1} + 24^{2}/2a
dividing by 18 and 24:
2.2 = t_{1} + 9/a
2.7 = t_{1} + 12/a
subtracting: 0.5 = 3/a, i.e. a = 6m.s^{2}
substituting: 2.7 = t_{1} + 12/6, i.e. t_{1} = 0.7s
