A pickup truck moving at 28 m/s must come to a stop in a distance of 191 meters to avoid hitting a boulder that has fallen onto the road. How much time does the driver have to avoid an accident?
How much time?
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Use the kinematic equation
Δx=½(vi+vf)t with vi=28 m/s, vf=0, and Δx=191 m,
and solve for t.
Hey, Sun -- ave speed of 14 m/s covers 191m in 10 +3 + 9/14 sec, or 13.6s <== this "ave speed" approach can be quite handy! Regards :)
V^2 = Vo^2 + 2ad
0 = 28^2 + 2a(191)
Solve for a,
a = -2.05 m/sec^2
V = Vo + at
0 = 28 - 2.05t
t = 13.6 sec
There is a shortcut formula for the uniformly accelerated motion, relating the distance traveled, initial and final velocities, and acceleration.
Here a is acceleration and vi and vf are initial and final velocities, respectively.
In your case vf=0 (truck stopped), vi=28 m/s, d=191 m.
191=(02-282)/(2*a) or 191=-392/a, from which a≈-2.05 m/s2. Now, when you know acceleration, time needed to stop can be computed from the formula:
a=(vf-vi)/t or t=(vf-vi)/a; Plugging in numbers gives t=-28/(-2.05)≈13.6 s.