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The figure below shows velocity time graph for a car in metres. a) Find the total distance covered by the car in metres. b) Calculate the deceleration of the car.

Physics

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Answers

2 parts
a

1500 m\text{1500 m}

b

12 m/s2\text{12 m/s}^2

Step-by-step solution

5 steps

Part a· 3 steps

  1. Gawanya grafu katika maeneo matatu.

    Grafu ya kasi-muda inaweza kugawanywa katika maeneo matatu ili kukokotoa umbali wote uliosafiriwa, kwani umbali ni eneo chini ya grafu ya kasi-muda.

    • Eneo la kwanza (0s hadi 5s) ni pembetatu.
    • Eneo la pili (5s hadi 25s) ni mstatili.
    • Eneo la tatu (25s hadi 30s) ni pembetatu.
  2. Kokotoa umbali kwa kila eneo.

    • Umbali wa kwanza (pembetatu):
    Umbali1=12×msingi×urefu=12×(5 s−0 s)×60 m/s=12×5 s×60 m/s=150 m\text{Umbali}_1 = \frac{1}{2} \times \text{msingi} \times \text{urefu} = \frac{1}{2} \times (5 \text{ s} - 0 \text{ s}) \times 60 \text{ m/s} = \frac{1}{2} \times 5 \text{ s} \times 60 \text{ m/s} = 150 \text{ m}
    • Umbali wa pili (mstatili):
    Umbali2=urefu×upana=(25 s−5 s)×60 m/s=20 s×60 m/s=1200 m\text{Umbali}_2 = \text{urefu} \times \text{upana} = (25 \text{ s} - 5 \text{ s}) \times 60 \text{ m/s} = 20 \text{ s} \times 60 \text{ m/s} = 1200 \text{ m}
    • Umbali wa tatu (pembetatu):
    Umbali3=12×msingi×urefu=12×(30 s−25 s)×60 m/s=12×5 s×60 m/s=150 m\text{Umbali}_3 = \frac{1}{2} \times \text{msingi} \times \text{urefu} = \frac{1}{2} \times (30 \text{ s} - 25 \text{ s}) \times 60 \text{ m/s} = \frac{1}{2} \times 5 \text{ s} \times 60 \text{ m/s} = 150 \text{ m}
  3. Jumlisha umbali wote.

    Jumla ya umbali=Umbali1+Umbali2+Umbali3=150 m+1200 m+150 m=1500 m\text{Jumla ya umbali} = \text{Umbali}_1 + \text{Umbali}_2 + \text{Umbali}_3 = 150 \text{ m} + 1200 \text{ m} + 150 \text{ m} = 1500 \text{ m}

    Jumla ya umbali uliosafiriwa na gari ni 1500 m\boxed{\text{1500 m}}.

Part b

12 m/s2\text{12 m/s}^2

  1. Tambua kipindi cha kupungua kasi.

    Kupungua kasi (deceleration) hutokea wakati kasi inapungua. Kwenye grafu, hii ni kati ya t=25 st = 25 \text{ s} na t=30 st = 30 \text{ s}.

  2. Kokotoa kupungua kasi.

    Kupungua kasi ni mteremko wa grafu katika kipindi hicho.

    Kupungua kasi=mabadiliko ya kasimabadiliko ya muda\text{Kupungua kasi} = \frac{\text{mabadiliko ya kasi}}{\text{mabadiliko ya muda}}

    Kasi ya awali (uu) = 60 m/s60 \text{ m/s} (kwa t=25 st = 25 \text{ s})
    Kasi ya mwisho (vv) = 0 m/s0 \text{ m/s} (kwa t=30 st = 30 \text{ s})
    Mabadiliko ya muda (Δt\Delta t) = 30 s−25 s=5 s30 \text{ s} - 25 \text{ s} = 5 \text{ s}

    Kupungua kasi=0 m/s−60 m/s5 s=−60 m/s5 s=−12 m/s2\text{Kupungua kasi} = \frac{0 \text{ m/s} - 60 \text{ m/s}}{5 \text{ s}} = \frac{-60 \text{ m/s}}{5 \text{ s}} = -12 \text{ m/s}^2

    Thamani hasi inaashiria kupungua kasi. Kwa hivyo, ukubwa wa kupungua kasi ni 12 m/s212 \text{ m/s}^2.
    Kupungua kasi ya gari ni 12 m/s2\boxed{\text{12 m/s}^2}.

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