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A train rides at 28 m.s-1 in a easterly direction on a straight track after which it brakes for 5 s and reaches a velocity of 15 m.s-1. Calculate the trains acceleration.

Physics

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Part 4.1· 3 steps

A truck starts from rest and reaches a velocity of 12 m⋅s−112 \text{ m}\cdot\text{s}^{-1} in 3 seconds. Calculate the acceleration of the truck if the original direction of motion is eastwards.

  • vi=0 m⋅s−1v_i = 0 \text{ m}\cdot\text{s}^{-1}
  • Δt=3 s\Delta t = 3 \text{ s}
  • vf=12 m⋅s−1v_f = 12 \text{ m}\cdot\text{s}^{-1}

  1. Write the formula for acceleration.

    a=vf−viΔta = \frac{v_f - v_i}{\Delta t}

  2. Substitute the given values into the formula.

    a=12 m⋅s−1−0 m⋅s−13 sa = \frac{12 \text{ m}\cdot\text{s}^{-1} - 0 \text{ m}\cdot\text{s}^{-1}}{3 \text{ s}}

  3. Calculate the acceleration.

    a=12 m⋅s−13 sa = \frac{12 \text{ m}\cdot\text{s}^{-1}}{3 \text{ s}} a=4 m⋅s−2 Easta = \boxed{4 \text{ m}\cdot\text{s}^{-2} \text{ East}}

Part 4.2· 3 steps

An object that moves at 12 m⋅s−112 \text{ m}\cdot\text{s}^{-1} north reaches a velocity of 53 m⋅s−153 \text{ m}\cdot\text{s}^{-1} after 5 seconds. Calculate the acceleration of the object.

  • vi=12 m⋅s−1v_i = 12 \text{ m}\cdot\text{s}^{-1}
  • Δt=5 s\Delta t = 5 \text{ s}
  • vf=53 m⋅s−1v_f = 53 \text{ m}\cdot\text{s}^{-1}

  1. Write the formula for acceleration.

    a=vf−viΔta = \frac{v_f - v_i}{\Delta t}

  2. Substitute the given values into the formula.

    a=53 m⋅s−1−12 m⋅s−15 sa = \frac{53 \text{ m}\cdot\text{s}^{-1} - 12 \text{ m}\cdot\text{s}^{-1}}{5 \text{ s}}

  3. Calculate the acceleration.

    a=41 m⋅s−15 sa = \frac{41 \text{ m}\cdot\text{s}^{-1}}{5 \text{ s}} a=8.2 m⋅s−2 Northa = \boxed{8.2 \text{ m}\cdot\text{s}^{-2} \text{ North}}

Part 4.3· 3 steps

What is the meaning of your answer in Question 3.1?

I cannot answer this question as Question 3.1 is not provided in the image or previous context.

5.1 A train rides at 28 m⋅s−128 \text{ m}\cdot\text{s}^{-1} in an easterly direction on a straight track after which it brakes for 5 s5 \text{ s} and reaches a velocity of 15 m⋅s−115 \text{ m}\cdot\text{s}^{-1}. Calculate the train's acceleration.

  • vi=28 m⋅s−1v_i = 28 \text{ m}\cdot\text{s}^{-1}
  • Δt=5 s\Delta t = 5 \text{ s}
  • vf=15 m⋅s−1v_f = 15 \text{ m}\cdot\text{s}^{-1}

  1. Write the formula for acceleration.

    a=vf−viΔta = \frac{v_f - v_i}{\Delta t}

  2. Substitute the given values into the formula.

    a=15 m⋅s−1−28 m⋅s−15 sa = \frac{15 \text{ m}\cdot\text{s}^{-1} - 28 \text{ m}\cdot\text{s}^{-1}}{5 \text{ s}}

  3. Calculate the acceleration.

    a=−13 m⋅s−15 sa = \frac{-13 \text{ m}\cdot\text{s}^{-1}}{5 \text{ s}} a=−2.6 m⋅s−2 East (or 2.6 m⋅s−2 West)a = \boxed{-2.6 \text{ m}\cdot\text{s}^{-2} \text{ East (or } 2.6 \text{ m}\cdot\text{s}^{-2} \text{ West)}}

Part 5.2· 3 steps

A motorcycle that moves at 30 m⋅s−130 \text{ m}\cdot\text{s}^{-1} north on a straight road, slams on brakes and stops after 6 s6 \text{ s}. Calculate the acceleration of the motorcycle.

  • vi=30 m⋅s−1v_i = 30 \text{ m}\cdot\text{s}^{-1}
  • Δt=6 s\Delta t = 6 \text{ s}
  • vf=0 m⋅s−1v_f = 0 \text{ m}\cdot\text{s}^{-1} (stops)

  1. Write the formula for acceleration.

    a=vf−viΔta = \frac{v_f - v_i}{\Delta t}

  2. Substitute the given values into the formula.

    a=0 m⋅s−1−30 m⋅s−16 sa = \frac{0 \text{ m}\cdot\text{s}^{-1} - 30 \text{ m}\cdot\text{s}^{-1}}{6 \text{ s}}

  3. Calculate the acceleration.

    a=−30 m⋅s−16 sa = \frac{-30 \text{ m}\cdot\text{s}^{-1}}{6 \text{ s}} a=−5 m⋅s−2 North (or 5 m⋅s−2 South)a = \boxed{-5 \text{ m}\cdot\text{s}^{-2} \text{ North (or } 5 \text{ m}\cdot\text{s}^{-2} \text{ South)}}
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