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    a cheetah can maintain a constant acceleration of 10.0 m/s2 for a time interval of 1.30 seconds. if the cheetah starts from rest, how long will it take the cheetah to move at 25.0 mph?

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    get a cheetah can maintain a constant acceleration of 10.0 m/s2 for a time interval of 1.30 seconds. if the cheetah starts from rest, how long will it take the cheetah to move at 25.0 mph? from EN Bilgi.

    Solved A cheetah can maintain a constant acceleration of

    Answer to Solved A cheetah can maintain a constant acceleration of

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    Solved A cheetah can maintain a constant acceleration of

    Source : www.chegg.com

    3.4 Motion with Constant Acceleration

    v0=10m/s,a=2.00m/s2v0=10m/s,a=2.00m/s2, and x = 200 m.

    We need to solve for t. The equation x=x0+v0t+12at2x=x0+v0t+12at2 works best because the only unknown in the equation is the variable t, for which we need to solve. From this insight we see that when we input the knowns into the equation, we end up with a quadratic equation.

    We need to rearrange the equation to solve for t, then substituting the knowns into the equation:

    200m=0m+(10.0m/s)t+12(2.00m/s2)t2.200m=0m+(10.0m/s)t+12(2.00m/s2)t2.

    We then simplify the equation. The units of meters cancel because they are in each term. We can get the units of seconds to cancel by taking t = t s, where t is the magnitude of time and s is the unit. Doing so leaves

    200=10t+t2.200=10t+t2. We then use the quadratic formula to solve for t,

    t2+10t200=0t=b±b24ac2a,t2+10t200=0t=b±b24ac2a, which yields two solutions: t = 10.0 and t = −20.0. A negative value for time is unreasonable, since it would mean the event happened 20 s before the motion began. We can discard that solution. Thus,

    t=10.0s.t=10.0s.

    3.4 Motion with Constant Acceleration

    Source : courses.lumenlearning.com

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