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    Neither the distance covered before the second brake or the time is given.. :confused:

    A car comes to a stop from a speed of 30 m/s in a distance of 830m. The driver brakes so as to produce a deceleration of 1/2 m/s^2 to begin with, and then brakes harder to produce a deceleration of 2/3 m/s^2. Find the speed of the car at the instant when the deceleration is increased, and the total time the car takes to stop..

    Any hints on how to solve this?
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    You're given the total distance, so let the distance covered during the first deceleration be S, then during the second part he covers a distance of (830-S).

    With the rest of the data you should be able to formulate two SUVAT expressions and solve (Hint: v^2=u^2+2as)
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    cheers
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    (Original post by marcusmerehay)
    You're given the total distance, so let the distance covered during the first deceleration be S, then during the second part he covers a distance of (830-S).

    With the rest of the data you should be able to formulate two SUVAT expressions and solve (Hint: v^2=u^2+2as)
    This.

    Also, you might have to look at the total change in velocity to produce another equation:

     v = a_1t_1 + a_2t_2
 
 
 
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Updated: March 19, 2011
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