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A driver starts a push car. Speed of 100km/h to a displacement of 88km at a constant acceleration.

Physics

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Answer

2.76 m/s2\text{2.76 m/s}^2

Step-by-step solution

3 steps

Here's the solution to the problem.

  1. Convert the initial and final velocities from kilometers per hour to meters per second.

    The conversion factor is 1000 m3600 s=518 m/s\frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}.
    Initial velocity v0=10 km/h=10×518 m/s=5018 m/s=259 m/sv_0 = 10 \text{ km/h} = 10 \times \frac{5}{18} \text{ m/s} = \frac{50}{18} \text{ m/s} = \frac{25}{9} \text{ m/s}.
    Final velocity v=80 km/h=80×518 m/s=40018 m/s=2009 m/sv = 80 \text{ km/h} = 80 \times \frac{5}{18} \text{ m/s} = \frac{400}{18} \text{ m/s} = \frac{200}{9} \text{ m/s}.

  2. Use the kinematic equation that relates initial velocity, final velocity, displacement, and constant acceleration.

    The relevant equation is v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x.
    We need to solve for acceleration aa. Rearranging the equation:

    a=v2−v022Δxa = \frac{v^2 - v_0^2}{2\Delta x}

  3. Substitute the values and calculate the acceleration.

    Given displacement Δx=88 m\Delta x = 88 \text{ m}.

    a=(2009 m/s)2−(259 m/s)22×88 ma = \frac{\left(\frac{200}{9} \text{ m/s}\right)^2 - \left(\frac{25}{9} \text{ m/s}\right)^2}{2 \times 88 \text{ m}} a=4000081 m2/s2−62581 m2/s2176 ma = \frac{\frac{40000}{81} \text{ m}^2/\text{s}^2 - \frac{625}{81} \text{ m}^2/\text{s}^2}{176 \text{ m}} a=40000−62581 m2/s2176 ma = \frac{\frac{40000 - 625}{81} \text{ m}^2/\text{s}^2}{176 \text{ m}} a=3937581 m2/s2176 ma = \frac{\frac{39375}{81} \text{ m}^2/\text{s}^2}{176 \text{ m}} a=3937581×176 m/s2a = \frac{39375}{81 \times 176} \text{ m/s}^2 a=3937514256 m/s2a = \frac{39375}{14256} \text{ m/s}^2 a≈2.762 m/s2a \approx 2.762 \text{ m/s}^2

    The constant acceleration is 2.76 m/s2∗\text{2.76 m/s}^2*.

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