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Thread: Kinematics: The Motion of Particles

  1. #1
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    Kinematics: The Motion of Particles

    A train takes 5 minutes to cover 3km between stations P and Q. Starting from rest at P, it accelerates at a constant rate to a speed of 40km/hr and maintains this speed for a while and then is brought uniformly to rest at Q. If the braking takes three times as long as the accelerate, find the time taken for the train to accelerate.

    I worked out that 3000=0*t+0.5at^2+(100/9)*t+0.5a*3t^2+100/9*16t applies to the answer. and t=15s, a= (20/27)ms^-2

    but how do you work t and a out?

    Million thanks!
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  2. #2
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    Hello BabyMilo
    Quote Originally Posted by BabyMilo View Post
    A train takes 5 minutes to cover 3km between stations P and Q. Starting from rest at P, it accelerates at a constant rate to a speed of 40km/hr and maintains this speed for a while and then is brought uniformly to rest at Q. If the braking takes three times as long as the accelerate, find the time taken for the train to accelerate.

    I worked out that 3000=0*t+0.5at^2+(100/9)*t+0.5a*3t^2+100/9*16t applies to the answer. and t=15s, a= (20/27)ms^-2

    but how do you work t and a out?

    Million thanks!
    Since the speed we're given is measured in km/hr, and the distance in km, let's suppose that the train accelerates for $\displaystyle t$ hr. So it brakes for $\displaystyle 3t$ hr and travels at constant speed for $\displaystyle (\tfrac{1}{12}-4t)$ hr, since $\displaystyle 5$ min $\displaystyle = \tfrac{1}{12}$ hr.

    Draw a velocity-time graph, which is in the form of a trapezium with horizontal, parallel sides of length $\displaystyle \tfrac{1}{12}$ (along the t-axis) and $\displaystyle (\tfrac{1}{12}-4t)$ along the line where $\displaystyle v = 40$.

    Now the area under this graph represents the distance travelled. So:

    $\displaystyle \tfrac12$ (sum of parallel sides) x (perpendicular distance between them) $\displaystyle =\tfrac12(\tfrac{1}{12}+\tfrac{1}{12}-4t)40=3$

    $\displaystyle \Rightarrow 20(\tfrac{1}{6}-4t)=3$

    $\displaystyle \Rightarrow 4t=\frac{1}{6}-\frac{3}{20}=\frac{1}{60}$

    $\displaystyle \Rightarrow t = \frac{1}{240}$ hr.

    So the acceleration time $\displaystyle = \tfrac14$ min $\displaystyle = 15$ seconds.

    Grandad
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