A car driver traveling on a linear segment of a level road at 60 miles per hour (mph) slows down to 40 miles per hour in a road construction zone. If the car traveled a distance of 360 ft from the original speed to the final speed then what is the car deceleration rate in ft per sec2

Explanation:

given

initial velocity u=60mph to ft per sec= 88ft/s

final velocity v=40 mph to ft per sec= 58.6 ft/s

distance s=360ft

convert feet to mile 360ft is 0.06miles

we can apply the expression

v^2=u^2-2as

58.6^2=88^2-2*a*360

3433.96=7744-720a

3433.96-7744=-720a

-4310.04=-720a

a=4310.04/720

a=5.98 ft/sec

protons in

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