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How many seconds is it to stop traveling 45 mph?

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That depends on the deceleration applied, in which case

time = 2x45 mph/a = 90/a

with a (the deceleration) measured in miles per hour per hour to give the time in hours.


If you mean the time that would be taken using the stopping distances in the UK Highway Code, then, assuming a constant deceleration (from the moment the brakes are applied after the thinking distance):

The stopping distance at 45 mph is 45 ft (thinking distance) + 101.25 ft (braking distance)

[braking distance is speed2 / 20 ft = 452 / 20 ft = 101.25 ft]

The thinking distance is travelled at 45 mph, giving:

think_time = 45 ft / 45 mph
= 45 ft /(45 x 5280 ft / 3600 seconds)
= 45 x 3600 / (45 x 5280) seconds
= 3600/5280 seconds
= 15/22 seconds
(This time is constant for all the emergency stopping distances given in the Highway Code.) Two equations of motion can be used to find the stopping time knowing the initial speed and distance (the final speed is zero):

final_velocity2 = initial_velocity2 + 2 x acceleration x distance
final_velocity = 0
→ acceleration = - initial_velocity2 / (2 x distance)
distance = initial_velocity x time + 1/2 x acceleration x time2
→ distance = initial_velocity x time - (1/4 x initial_velocity2 / distance) x time2
→ time2 - (4 x distance / initial_velocity) x time + 4 x distance2 / initial_velocity2 = 0
→ (time - 2 x distance/initial_velocity)2 = 0
→ breaking_time = 2 x distance / initial_velocity
= 2 x (101.25 ft) / (45 x 5280 / 3600 ft per sec)
= 202.5 x 3600 / (45 x 5280) seconds
= 9x15/2x22
→ total_stopping_time = 15/22 + 9x15/2x22 seconds
= 11x15/2x22 seconds
= 15/4 seconds
= 33/4 seconds
= 3.75 seconds.

Thanks for the feedback!

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