The \"reaction time\" of the average automobile driver is about 0.7 s . (The rea

Question

The "reaction time" of the average automobile driver is about 0.7 s . (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) An automobile can slow down with an acceleration of 13.8 ft/s2

a) Compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 13.0 mi/h . (in a school zone)

b) Compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 55.8 mi/h .

Explanation / Answer

Given :-

Reaction Time = 0.7 s

Acceleration = 13.8 ft/s2

A) initial velocity (u) = 13 mi/h

first need to convert mi/h to ft/s

As We know,

There are 5280 feet in 1 mile

so 13 mi/h = (13 * 5280)/ 3600 sec

19.06 ft/s

Now from equation of motion,

v^2 = u^2 + (-2as) {-ve sign due to acceleration is in opposite to the direction of velocity}

Where v is final velocity =0 because automobile will stop after break

u = initial velocity = 19.06 ft/s

a =accleration =13.8 ft / s^2

Now by putting all these value in above equation we get

0 = (19.06)^2 - 2 * 13.8 * s

s= 13.16 ft

Now distance travel during reaction time = initial velocity * reaction tine

= 19.06 * 0.7 ft

=13.342 ft

total distance = 13.342ft + 13.16ft = 26.502 -------------------------------------(i) ans

B) initial velocity (u) = 55.8 mi/h

first need to convert mi/h to ft/s

As We know,

There are 5280 feet in 1 mile

so 55.8 mi/h = (55.8 * 5280)/ 3600 sec

81.84 ft/s

Now from equation of motion,

v^2 = u^2 + (-2as) {-ve sign due to acceleration is in opposite to the direction of velocity}

Where v is final velocity =0 because automobile will stop after break

u = initial velocity = 81.84 ft/s

a =accleration =13.8 ft / s^2

Now by putting all these value in above equation we get

0 = (81.84)^2 - 2 * 13.8 * s

s= 242.67 ft

Now distance travel during reaction time = initial velocity * reaction tine

= 81.84 * 0.7 ft

=57.288 ft

total distance = 242.67ft + 57.288ft = 300ft -------------------------------------(ii) ans

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