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/sec^{2}.

B) Distance traveled formula d = v_{i}t_{ } + 1/2 at^{2} where v_{i
}is the initial velocity, a is the acceleration and t is the time.

v_{i} = 632 mi/hr = 926.933 ft/sec

a = -662.095 ft/sec^{2}

t = 1.4 sec

d=(926.933 ft/sec)(1.4 sec) + 1/2(-662.095 ft/sec^{2})(1.96 sec^{2}) = 648.8531 ft

C) Done above

## Comments