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... time = (25 ms-1 + 0 ms⁻¹)2 x 30 s = 12.5 ms⁻¹ x 30 s = 375 m$

Mathematics

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Answer

21.25 ms−1\text{21.25 ms}^{-1}

8000 J\text{8000 J}

Step-by-step solution

6 steps

... time = \frac{(25 ms^{-1} + 0 ms^{-1})}{2} \times 30 s = 12.5 ms^{-1} \times 30 s = 375 \text{ m}$

  1. Calculate the total distance traveled.

    Dtotal=d1+d2+d3+d4=900 m+600 m+4500 m+375 m=6375 mD_{\text{total}} = d_1 + d_2 + d_3 + d_4 = 900 \text{ m} + 600 \text{ m} + 4500 \text{ m} + 375 \text{ m} = 6375 \text{ m}
  2. Calculate the total time taken.

    Ttotal=60 s+30 s+180 s+30 s=300 sT_{\text{total}} = 60 \text{ s} + 30 \text{ s} + 180 \text{ s} + 30 \text{ s} = 300 \text{ s}
  3. Calculate the average speed.

    Average speed=DtotalTtotal=6375 m300 s=21.25 ms−1\text{Average speed} = \frac{D_{\text{total}}}{T_{\text{total}}} = \frac{6375 \text{ m}}{300 \text{ s}} = 21.25 \text{ ms}^{-1}

    The average speed of the bus is 21.25 ms−1\boxed{\text{21.25 ms}^{-1}}.

    c) Estimate the heat gained by the brake pad.

  4. Identify the given values.

    • Mass of brake pad (mm) = 200 g=0.2 kg200 \text{ g} = 0.2 \text{ kg}
    • Initial temperature (T1T_1) = 40∘C40^\circ\text{C}
    • Final temperature (T2T_2) = 80∘C80^\circ\text{C}
    • Specific heat capacity (cc) = 1000 Jkg−1K−11000 \text{ Jkg}^{-1}\text{K}^{-1}
  5. Calculate the change in temperature (ΔT\Delta T).

    ΔT=T2−T1=80∘C−40∘C=40∘C\Delta T = T_2 - T_1 = 80^\circ\text{C} - 40^\circ\text{C} = 40^\circ\text{C}

    Since a change of 1∘C1^\circ\text{C} is equal to a change of 1 K1 \text{ K}, ΔT=40 K\Delta T = 40 \text{ K}.

  6. Calculate the heat gained (QQ) using the formula Q=mcΔTQ = mc\Delta T.

    Q=(0.2 kg)×(1000 Jkg−1K−1)×(40 K)Q = (0.2 \text{ kg}) \times (1000 \text{ Jkg}^{-1}\text{K}^{-1}) \times (40 \text{ K}) Q=8000 JQ = 8000 \text{ J}

    The heat gained by the brake pad is 8000 J\boxed{\text{8000 J}}.

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