Lesson 5, Topic 4
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# Combining the Velocity-Time Graphs

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The total distance travelled is the total area under the graph from the graph, the shape is that of a trapezium.

∴ Total distance covered = Area of a trapezium

=   $$\frac{1}{2}$$(Sum of parallel sides) x h

=   $$\frac{1}{2} \scriptsize (a \: + \: b) \: \times \: h$$

Also, breaking the graph into components, S1, S2, S3

S1 is a right angle triangle = distance covered = Area of a right angle triangle

S1 = $$\frac{1}{2} \scriptsize \: \times \: base \: \times \: height \\ = \frac{1}{2} \scriptsize bh$$

S2 = Area of a rectangle = total distance covered = L x b

S3 = Area of a right angle triangle = $$\frac{1}{2} \scriptsize \: \times \: base \: \: \times \: height \\ =\frac{1}{2} \scriptsize bh$$

∴ the total distance covered = S1 S2 + S3

= $$\normalsize \frac{1}{2} \scriptsize bh \: + \: (L \: \times \: b) \: + \: \normalsize \frac{1}{2} \scriptsize bh$$

Examples:

i. A car is travelling with initial velocity of 30ms-1 and attains a final velocity of 80ms-1 in 20seconds. Find the acceleration of the car.

Solution: u = 30ms-1, v = 80ms-1, t = 20seconds

a = $$\frac {v \: – \: u}{t} \\ = \frac {80\: – \: 30}{20} \\ = \frac {50}{20} \\= \scriptsize 2.5ms^{-2}$$

ii. A Toyota car started from rest and accelerated to 4.0 m/s for 2 seconds. The car then moved with uniform velocity for 4 seconds. When the brakes were applied, the car slowed down to a stop for 2 seconds.

Calculate: (a) the acceleration of the car. (b) the deceleration of the car. (c) the distance travelled by car in each phase of the motion.  (d) the total distance travelled.

a. When the car accelerated from rest to 4.0 m/s, u = 0 m/s, v = 4.0 m/s, time, t = 2 seconds.

acceleration = $$\frac {v \: – \: u}{t} \\ = \frac {4 \: – \: 0}{2}$$

acceleration = 2ms-2

b. When the brakes were applied, the car decelerated ( i.e slowed down) from 4.0 m/s to 0 m/s for 2 seconds.

Then, u = 1.0 m/s, v = 0 m/s, t = 2 seconds.

deceleration = $$\frac {v \: – \: u}{t} \\ = \frac {0 \: – \: 4}{2}$$

acceleration = – 2ms-2

c. In phase 1 (acceleration stage)

Distance covered = Area of a right angle triangle

Distance covered = $$\frac{1}{2} \scriptsize\: \times \: b \: \times \: h \\ = \frac{1}{2}\scriptsize\: \times \: 2s \: \times \: 4ms^{-1} \\ \scriptsize = 4 m$$

In phase 2 (uniform velocity stage)

Area of a square = total distance covered = l x b = 4m/s x 4s = 16m

In phase 3 (deceleration stage)

Distance covered = Area of a right angle triangle

Distance covered = $$\frac{1}{2} \scriptsize \: \times \: b \: \times \: h \\ = \frac{1}{2} \scriptsize\: \times \: 2s \: \times \: 4ms^{-1} \\ \scriptsize = 4 m$$

d. The total distance travelled = phase 1 + phase 3 + phase 3

= 4m + 16m + 4m = 24m

OR

Total distance covered = Area of a trapezium

:- $$\frac{1}{2} \scriptsize\: \times \: (a \: + \: b) \: \times \: h \\ = \frac{1}{2} \scriptsize \: \times \: (4 \: + \: 8) \: \times \: 4 \\ \scriptsize = 6 \: \times \: 4 \\ \scriptsize = 24m$$

Evaluation Questions

1. Define (i) Velocity.

(ii) Acceleration

2. A body accelerates from rest and moves with a uniform acceleration of 10m/s2. What distance does it cover in the last one second of its motion?

3. A car is brought to rest from a speed of 25m/s in 10 seconds. Find the retardation.

4. A mango dropped from a height of 100m above the ground. Calculate the velocity of the mango just before hitting the ground.

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