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If a man rides a rocket-propelled sled that moved down the track at 632mi/h. He and the sled were brought to rest in 1.4s.

A. How do you determine the deceleration he experienced?
B. What was the distanced he traveled during this deceleration?
C. How do you finish or complete the conversion?
632 mi x 5,280ft  x    1hr     x 1min.   x ____ft.
     hr.       1 mile     60min      60sec.            s

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Jim S. | Physics (and math) are fun, reallyPhysics (and math) are fun, really
4.7 4.7 (186 lesson ratings) (186)
A. a=ΔV/Δt where ΔV=0-931.3 = -931.3 ft/sec and Δt=1.4 sec (using 60mph=88ft/sec)
     so a=-931.3/1.4 = -665.2 ft/sec2  is the average acceleration of the sled during the slowdown
B. D= Vt + 1/2 at= 931.3x1.4 - .5x665.2x1.42 = 651.9 ft
C.There is nothing to compete the four factors you have converts 632 mph to ft/sec the exact number is 926.9 ft/sec


Not sure where you got 931.3 ft/sec for 632 mph otherwise our methods are the same.  Just pointing this out so that Dana doesn't get confused with our slightly different answers.
David S. | Wise Math TutorWise Math Tutor
5.0 5.0 (58 lesson ratings) (58)
Hi Dana,
A) Convert 632 mi/hr to ft/sec by using the conversion in your question above(note small typo in what you typed) then divide by 1.4 sec to get your acceleration.  It will be minus since it is deceleration. Your conversion above has one small typo.  It should read 632 mi/hr. times 5,280 ft/min times 1 hr/60 min times 1 min/60 sec = — ft/sec not times — ft/sec.  The acceleration will then be -926.33 ft/sec divided by 1.4 sec = -662.095 ft/sec2.
B) Distance traveled formula d = vit    + 1/2 at2 where vi is the initial velocity, a is the acceleration and t is the time.
vi =  632 mi/hr = 926.933 ft/sec
a = -662.095 ft/sec2
t = 1.4 sec
d=(926.933 ft/sec)(1.4 sec) + 1/2(-662.095 ft/sec2)(1.96 sec2) = 648.8531 ft
C) Done above