Question #171115

at the end of the race, a runner decelerates from a velocity of 9m/s^2. how far does she travel in next 5s?




Expert's answer

I assume, the initial velocity was v0=9m/sv_0 = 9m/s. Then the acceleration was (by definition):


a=vfv0t=v0ta = \dfrac{v_f-v_0}{t} = -\dfrac{v_0}{t}

where vf=0v_f = 0, is the final speed of the runner, t=5st = 5s is the time she decelerated.

The distance travelled in this time was:


d=v0tat22=v0tv0t2=v0t2d = v_0t - \dfrac{at^2}{2} = v_0t - \dfrac{v_0t}{2} = \dfrac{v_0t}{2}

Thus, obtain:


d=9m/s5s2=22.5md = \dfrac{9m/s\cdot 5s}{2} = 22.5m

Answer. 22.5 m.


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