How much time?

Question

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?

Answer 1

Use the kinematic equation

 

Δx=½(v_i+v_f)t with v_i=28 m/s, v_f=0, and Δx=191 m,

 

and solve for t.

Answer 2

 

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 :) 

Answer 3

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

Answer 4

There is a shortcut formula for the uniformly accelerated motion, relating the distance traveled, initial  and final velocities, and acceleration.

 

d=(v_f²-v_i²)/(2a).

 

Here a is acceleration and v_i and v_f are initial and final velocities, respectively.

 

In your case v_f=0 (truck stopped), v_i=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=(v_f-v_i)/t or t=(v_f-v_i)/a; Plugging in numbers gives t=-28/(-2.05)≈13.6 s.