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How much time?

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?

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Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
1
Use the kinematic equation
 
Δx=½(vi+vf)t with vi=28 m/s, vf=0, and Δx=191 m,
 
and solve for t.
Brad M. | Professional Tutor: Ultra Streamlined Math - Physics - AccountingProfessional Tutor: Ultra Streamlined M...
4.9 4.9 (218 lesson ratings) (218)
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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 :) 
Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
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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
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
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There is a shortcut formula for the uniformly accelerated motion, relating the distance traveled, initial  and final velocities, and acceleration.
 
d=(vf2-vi2)/(2a).
 
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.