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AJTurnbull Group Title

a car is traveling in a straight line and decelerating at a constant rate. after 3.2 s it traveled 88m and slowed to a speed of 17m/s what was the initial speed of the car?

  • 2 years ago
  • 2 years ago

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  1. nikvist Group Title
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    \[t=3.2s\quad,\quad s=88m\quad,\quad v=17m/s\]\[v=v_0-at\quad\Rightarrow\quad a=\frac{v_0-v}{t}\]\[v^2=v_0^2-2as=v_0^2-2\frac{v_0-v}{t}s=v_0^2-\frac{2s}{t}v_0+\frac{2s}{t}v\]\[v_0^2-\frac{2s}{t}v_0+\frac{2s}{t}v-v^2=0\]\[v_0=\frac{\frac{2s}{t}\pm\sqrt{\left(\frac{2s}{t}\right)^2-4\left(\frac{2s}{t}v-v^2\right)}}{2}=\]\[=\frac{s}{t}\pm\sqrt{\left(\frac{s}{t}\right)^2-\frac{2s}{t}v+v^2}=\frac{s}{t}\pm\sqrt{\left(\frac{s}{t}-v\right)^2}=\]\[=\frac{s}{t}\pm\left(\frac{s}{t}-v\right)\quad\mbox{(only sign +)}\]\[v_0=\frac{2s}{t}-v=38m/s\]

    • 2 years ago
  2. shubham Group Title
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    OR Use this equation of motion S = {(u+v)/2}t 88m = {(u+17)/2}3.2 u=55-13 = 38m/s

    • 2 years ago
  3. AJTurnbull Group Title
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    how did you get a?

    • 2 years ago
  4. shubham Group Title
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    For constant accelarations, velocity is linear function of time and hence you can assume mean of initial and final velocities to be the average velocity. So, you don't need a actually.

    • 2 years ago
  5. AJTurnbull Group Title
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    oh that makes it easy been trying to figure that out all night

    • 2 years ago
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