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A) i) Find the total distance covered by the car in metres. ii) Calculate the deceleration of the car.

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2 parts
ii

12 m/s2\text{12 m/s}^2

b

10:00 am\text{10:00 am}

Step-by-step solution

8 steps

20.a) i) Find the total distance covered by the car in metres.

The total distance covered is the area under the velocity-time graph. The graph is a trapezium.

Step 1: Identify the shape and its dimensions.
The maximum velocity is 60 m/s60 \, \text{m/s}.
The total time duration is 30 s30 \, \text{s}.
The duration of constant velocity is from t=5 st=5 \, \text{s} to t=25 st=25 \, \text{s}, which is 25−5=20 s25 - 5 = 20 \, \text{s}.

Step 2: Calculate the area of the trapezium.
The formula for the area of a trapezium is A=12(a+b)hA = \frac{1}{2}(a+b)h, where aa and bb are the lengths of the parallel sides and hh is the perpendicular height.
Here, a=30 sa = 30 \, \text{s}, b=20 sb = 20 \, \text{s}, and h=60 m/sh = 60 \, \text{m/s}.

Distance=12(30 s+20 s)×60 m/s\text{Distance} = \frac{1}{2} (30 \, \text{s} + 20 \, \text{s}) \times 60 \, \text{m/s} Distance=12(50 s)×60 m/s\text{Distance} = \frac{1}{2} (50 \, \text{s}) \times 60 \, \text{m/s} Distance=25 s×60 m/s\text{Distance} = 25 \, \text{s} \times 60 \, \text{m/s} Distance=1500 m\text{Distance} = 1500 \, \text{m}

The total distance covered by the car is 1500 m\boxed{\text{1500 m}}.

Part ii· 4 steps

Calculate the deceleration of the car.

Deceleration is the negative of the acceleration, which is the gradient of the velocity-time graph during the slowing down phase.

  1. Identify the segment of the graph where deceleration occurs.

    Deceleration occurs from t=25 st=25 \, \text{s} to t=30 st=30 \, \text{s}.

  2. Determine the initial and final velocities and the time interval for this segment.

    Initial velocity (uu) at t=25 st=25 \, \text{s} is 60 m/s60 \, \text{m/s}.
    Final velocity (vv) at t=30 st=30 \, \text{s} is 0 m/s0 \, \text{m/s}.
    Time interval (Δt\Delta t) is 30 s−25 s=5 s30 \, \text{s} - 25 \, \text{s} = 5 \, \text{s}.

  3. Calculate the acceleration.

    a=v−uΔta = \frac{v - u}{\Delta t} a=0 m/s−60 m/s5 sa = \frac{0 \, \text{m/s} - 60 \, \text{m/s}}{5 \, \text{s}} a=−60 m/s5 sa = \frac{-60 \, \text{m/s}}{5 \, \text{s}} a=−12 m/s2a = -12 \, \text{m/s}^2

  4. State the deceleration.

    Deceleration is the magnitude of the negative acceleration.

    Deceleration=∣−12 m/s2∣=12 m/s2\text{Deceleration} = |-12 \, \text{m/s}^2| = 12 \, \text{m/s}^2

    The deceleration of the car is 12 m/s2\boxed{\text{12 m/s}^2}.

Part b

10:00 am\text{10:00 am}

Determine the time of the day when the matatu caught up with the lorry.

  1. Define variables and set up equations for the distance covered by the lorry.

    Let tt be the time in hours after 8:00 am when the matatu catches up with the lorry.
    Lorry's speed (VLV_L) = 72 km/h72 \, \text{km/h}.
    Lorry's travel time = tt hours.
    Distance covered by lorry (DLD_L) = VL×t=72t kmV_L \times t = 72t \, \text{km}.

  2. Set up equations for the distance covered by the matatu.

    Matatu's speed (VMV_M) = 96 km/h96 \, \text{km/h}.
    Matatu started at 8:30 am, which is 0.5 hours after the lorry.
    Matatu's travel time = (t−0.5)(t - 0.5) hours.
    Distance covered by matatu (DMD_M) = VM×(t−0.5)=96(t−0.5) kmV_M \times (t - 0.5) = 96(t - 0.5) \, \text{km}.

  3. Equate the distances and solve for tt.

    When the matatu catches up with the lorry, their distances from Kisumu are equal:

    DL=DMD_L = D_M 72t=96(t−0.5)72t = 96(t - 0.5) 72t=96t−4872t = 96t - 48 48=96t−72t48 = 96t - 72t 48=24t48 = 24t t=4824t = \frac{48}{24} t=2 hourst = 2 \, \text{hours}

  4. Calculate the time of day.

    The matatu catches up with the lorry 2 hours after 8:00 am.
    Time of day = 8:00 am + 2 hours = 10:00 am.

    The matatu caught up with the lorry at 10:00 am\boxed{\text{10:00 am}}.

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