A particle P is moving in a straight line. The velocity-time graph(shown below) illustrates how the speed of P varies with time in a particular direction. Find the acceleration of the particle in each section.

The particle is at rest at point $O$.
For the first $5$ seconds, from $O$ to $A$, the particle moves forward.
Gradient of $OA = \frac{15}{5} = 3$
Acceleration $(𝑎) = \frac{𝑣}{𝑡} = \frac{15 𝑚𝑠^{−1}}{5 𝑠}$
$= 3 𝑚𝑠^{−2}$

For the next $2. 5$ seconds, from $A$ to $B$, the horizontal line shows that the velocity is constant.
Gradient of $AB = \frac{0}{2.5} = 0$
Acceleration $(𝑎) = \frac{0 𝑚𝑠^{−1}}{2.5 𝑠}$
$= 0 𝑚𝑠^{−2}$

It means there is no acceleration in this section.
For the next $7. 5$ seconds, from $B$ to $C$, the particle decelerates.
Gradient of $BC = \frac{−15}{7.5} =− 2$
Acceleration $(𝑎) = \frac{−15 𝑚𝑠^{−1}}{7.5 𝑠}$
$=− 2 𝑚𝑠^{−2}$
or, Deceleration $= 2 𝑚𝑠^{−2}$

Form the graph,
When $𝑡 = 15$, the object $P$ comes into rest and stops at point $C$, which is also called the stationary point.

For the next $2. 5$ seconds, the particle moves backward.
Gradient of $CD = \frac{−5}{2.5} =− 2$
Acceleration $(𝑎) = \frac{−5 𝑚𝑠^{−1}}{2.5 𝑠}$
$=− 2 𝑚𝑠^{−2}$

For the last $2. 5$ seconds, the particle again moves forward,
Gradient of $BC = \frac{5}{2.5} = 2$
Acceleration $(𝑎) = \frac{5 𝑚𝑠^{−1}}{2.5 𝑠}$
$= 2 𝑚𝑠^{−2}$
When $𝑡 = 20$, the object $P$ comes into rest again.
A lift moves upwards from rest and accelerates at $0. 9 𝑚𝑠^{−2}$ for $3$ seconds. The lift then travels for $6$ seconds at a constant speed and finally slows down with a constant deceleration, stopping in a further $4$ seconds.
Sketch the velocity-time graph. Find distance travelled and displacement by the lift.

The lift moves upwards from rest.
$⇒$ Plot $(0, 0)$
For the first $3$ seconds,
Acceleration $(𝑎) = 0. 9 𝑚𝑠^{−2}$
Velocity $=$ Acceleration $×$ Time
$= 0. 9 × 3$
$= 2. 7 𝑚𝑠^{−1}$
$⇒$ Plot $(3, 2. 7)$
For the next $6$ seconds,
The speed of the lift is constant, so the velocity does not change until $𝑡 = 9$ seconds.
$⇒$ Plot $(9, 2. 7)$
After that, the speed of the lift slows down with a constant deceleration and comes to rest in a further $4$ seconds.
$⇒$ Plot $(13, 0)$

Now, connect each points to draw velocity-time graph,

Total distance travelled by the lift $=$ Area of trapezium
$= \frac12 \bigg(𝑃_1 + 𝑃_2\bigg) × ℎ$
$= \frac12 \bigg(6 + 13\bigg) × 2.7$
$= \frac12 × 19 × 2. 7$
$= 25. 65$ metres

The motion of the lift was made straight upwards. So, for this case,
Distance $=$ Displacement $= 25. 65$ metres
Calculate the total distance and displacement of the object, which is represented by the velocity-time graph below.

The first is a right-angled triangle.
So, Area $= \frac12 × 5 × 15$
$= 37. 5$

The second part is a rectangle.
So, Area $= 2. 5 × 15$
$= 37. 5$

Since the third part is also a right-angled triangle.
So, Area $= \frac12 × 7. 5 × 15$
$= 56. 25$

Finally, we can split the final triangle into two right-angled triangles.
So calculating the areas of $4𝑡ℎ$ and $5𝑡ℎ$ triangle,
Area of each triangle $= \frac12 × 2. 5 × 5$
$= 6. 5$

Total distance $=$ Total area under the velocity-time graph.
$= 37. 5 + 37. 5 + 56. 25 + 6. 25 + 6. 25$
$= 143. 75$ metres

For the first $15$ seconds, the particle has moved forward in its motion.
But, in between $15$ seconds and $20$ seconds, the particle moves backward.
Also the area of the triangle is $= \frac12 × (− 5) × 5 =− 12. 5$.
Hence, the total displacement is $25$ less than the total distance.

Total distance $= 37. 5 + 37. 5 + 56. 25 − 12. 5$
$= 118. 75$ metres

4.12 - Velocity-time Graphs

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