You are driving your car along a country road at a speed of 25.2 m/s. As you come over the crest of a hill, you notice a farm tractor 41.0 m ahead of you on the road, moving in the same direction as you at a speed of 10.0 m/s. You immediately slam on your brakes and slow down with a constant acceleration of magnitude 7.00 m/s^2.
(a) Will you hit the tractor before you stop?
(b) How far will you travel before you stop or collide with the tractor?
(c) If you stop, how far is the tractor in front of you when you finally stop?
I've tried this problem over and over but I can't figure it out. If someone could show me how to do even one part of this problem that would be great. I just don't know what approach to use to solve it. Thanks so much!
(a) Will you hit the tractor before you stop?
(b) How far will you travel before you stop or collide with the tractor?
(c) If you stop, how far is the tractor in front of you when you finally stop?
I've tried this problem over and over but I can't figure it out. If someone could show me how to do even one part of this problem that would be great. I just don't know what approach to use to solve it. Thanks so much!
-
you: v initial = 25.2 m/s
slam brakes to a stop, use:
(v_final)^2 - (v_initial)^2 = 2*a*d
d= (-635.04)/(2* -7) = 45.4 m
How long does it take for you to stop?
45.4 m = (25.2)t - 1/2(7)t^2
have to use quadratic ...
t=3.6s
farm tractor was 41m ahead at t=0. At t=3.6s, the farm tractor has moved an additional (10m)*(3.6s) = 36m.
(a) No
(b) 45.4 m
(c) (36+41)m -45.4m = 31.6m or 32m
***************************************…
You don't *really* have to use the quadratic if you just consider how many seconds it takes to go from 25.2 m/s to zero if you decelerate at -7m/s^2 (it's obviously 3.6s), but official "solutions" to textbook problems will grind through the formal solution.
slam brakes to a stop, use:
(v_final)^2 - (v_initial)^2 = 2*a*d
d= (-635.04)/(2* -7) = 45.4 m
How long does it take for you to stop?
45.4 m = (25.2)t - 1/2(7)t^2
have to use quadratic ...
t=3.6s
farm tractor was 41m ahead at t=0. At t=3.6s, the farm tractor has moved an additional (10m)*(3.6s) = 36m.
(a) No
(b) 45.4 m
(c) (36+41)m -45.4m = 31.6m or 32m
***************************************…
You don't *really* have to use the quadratic if you just consider how many seconds it takes to go from 25.2 m/s to zero if you decelerate at -7m/s^2 (it's obviously 3.6s), but official "solutions" to textbook problems will grind through the formal solution.