MATHWITHCYE SPM KSSM Mathematics

Chapter 7: Graphs of Motion (Graf Gerakan)

Focus: Distance-Time & Speed-Time Graphs, Acceleration, Area as Distance & Average Speed

Total Marks
/ 30 Marks
Suggested Time: 45 Mins

Section A • Bahagian A [12 Marks]

Question 1 (Distance-Time Interpretation) [4 Marks]

A car travels from Town $P$ to Town $Q$. The distance-time graph shows the journey where the car travels $80\text{ km}$ in the first 1 hour, stops for 30 minutes at a rest area, and travels the remaining $60\text{ km}$ in 45 minutes.

(a) State the duration, in minutes, that the car was stationary. [1 mark]
(b) Calculate the speed, in $\text{km/h}$, of the car during the final 45 minutes. [2 marks]
(c) Calculate the average speed, in $\text{km/h}$, for the entire journey. [1 mark]

(a) Stationary time

30 minutes (or 0.5 hour)
P1

(b) Speed = $\frac{60\text{ km}}{45/60\text{ h}} = \frac{60}{0.75}$

$$= \mathbf{80}\text{ km/h}$$
K1 N1

(c) Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{80 + 60}{1 + 0.5 + 0.75} = \frac{140}{2.25}$

$$= \mathbf{62.22}\text{ km/h}$$
N1
Question 2 (Speed-Time Acceleration & Deceleration) [4 Marks]

The speed of an electric train increases uniformly from $15\text{ m/s}$ to $35\text{ m/s}$ in $8\text{ seconds}$, then maintains constant speed for $12\text{ seconds}$, and finally decelerates to a stop in $10\text{ seconds}$.

(a) Calculate the rate of change of speed (acceleration) during the first 8 seconds. [2 marks]
(b) Calculate the magnitude of deceleration during the final 10 seconds. [2 marks]

(a) Acceleration $a = \frac{35 - 15}{8} = \frac{20}{8}$

$$= \mathbf{2.5}\text{ m/s}^2$$
K1 N1

(b) Deceleration $= \frac{0 - 35}{10} = -3.5\text{ m/s}^2$

$$\text{Deceleration magnitude} = \mathbf{3.5}\text{ m/s}^2$$
K1 N1
Question 3 [4 Marks]

In a speed-time graph, a particle moves from rest to a speed of $v\text{ m/s}$ in $5\text{ seconds}$. The area under the graph for the first 5 seconds is $75\text{ m}$. Calculate the value of $v$.

Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height} = 75$

$$\frac{1}{2} \times 5 \times v = 75 \implies 2.5v = 75$$
K2

Solve for $v$

$$v = \mathbf{30}\text{ m/s}$$
N2

Section B • Bahagian B (HOTS / KBAT) [18 Marks]

Question 4 (Comprehensive Speed-Time Section B) [9 Marks]

The speed-time graph shows the motion of an express bus for a period of $T\text{ seconds}$. The bus starts from rest, accelerates uniformly to $24\text{ m/s}$ in $10\text{ seconds}$, cruises at this speed for $15\text{ seconds}$, and then decelerates to a stop in $(T - 25)\text{ seconds}$. The total distance travelled is $540\text{ m}$.

(a) State the uniform speed of the bus. [1 mark]
(b) Calculate the value of $T$. [4 marks]
(c) Calculate the deceleration of the bus during the final $(T - 25)\text{ seconds}$. [2 marks]
(d) Find the average speed of the bus for the whole journey. [2 marks]

(a) Uniform speed

$24\text{ m/s}$
P1

(b) Area of trapezium $= \frac{1}{2}(15 + T) \times 24 = 540$

$12(15 + T) = 540 \implies 15 + T = 45$
$$T = \mathbf{30}\text{ seconds}$$
K2 N2

(c) Final stage time $= 30 - 25 = 5\text{ s}$. Deceleration:

$$\text{Deceleration} = \frac{24 - 0}{5} = \mathbf{4.8}\text{ m/s}^2$$
K1 N1

(d) Average speed

$$\text{Average Speed} = \frac{540\text{ m}}{30\text{ s}} = \mathbf{18}\text{ m/s}$$
K1 N1
Question 5 (Dual-Vehicle Meeting Problem) [9 Marks]

Motorcyclist $A$ and Cyclist $B$ start simultaneously from the same point along a straight track. Motorcyclist $A$ moves with uniform acceleration from rest to $30\text{ m/s}$ in $12\text{ seconds}$ and maintains that speed. Cyclist $B$ travels with a constant uniform speed of $18\text{ m/s}$.

(a) Find the distance travelled by Motorcyclist $A$ during the first 12 seconds. [2 marks]
(b) Calculate the distance between Motorcyclist $A$ and Cyclist $B$ at $t = 12\text{ seconds}$. [3 marks]
(c) Calculate the time $t$, in seconds, when Motorcyclist $A$ overtakes Cyclist $B$. [4 marks]

(a) Distance of $A$ at 12s: Area of triangle $= \frac{1}{2}(12)(30)$

$$= \mathbf{180}\text{ m}$$
K1 N1

(b) Distance of $B$ at 12s $= 18 \times 12 = 216\text{ m}$.

$$\text{Distance between them} = 216 - 180 = \mathbf{36}\text{ m} \quad (B \text{ is ahead by } 36\text{ m})$$
K1 N2

(c) Let time after 12s be $t_1$. At overtake: $180 + 30t_1 = 18(12 + t_1)$

$180 + 30t_1 = 216 + 18t_1 \implies 12t_1 = 36 \implies t_1 = 3\text{ s}$
$$\text{Total time } t = 12 + 3 = \mathbf{15}\text{ seconds}$$
K2 N2
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