A San Francisco cable car is pulled by a cable running under the street. The cab

Question

A San Francisco cable car is pulled by a cable running under the street. The cable moves at a constant speed of 9.50 mi/hr. The person who operates the cable car (called a gripman) operates a grip that grabs hold of the moving cable. (See Figure 2.) The gripman can allow the cable to slide through the grip so that the car can move at a speed that is slower than the cable. In Figure 2 the cable is moving to the right. (Take this as the direction of the +x axis.) able Figure 2 Grip A cable car starts from rest as the gripman slowly tightens the grip on the cable, acquiring the speed of the cable in 50.0 ft. The cable car continues at this constant speed until the gripman sees tourists at the next car stop. At this point the gripman releases the cable and applies the brakes, stopping the cable car in 5.00 s ft between the stops. Convert 9.50 mi/hr to ft/s. (Hint: Use the conversion factor 60 mi/hr = 88 ft/s) (5 points) b. a. What is the cable car's acceleration during the first 50.0 ft of its motion? (Assume the acceleration is constant.) (5 points) c. How long did it take the cable car to reach the speed of the cable? (5 points) d. What is the cable car's acceleration when it comes to a stop? (Watch sign e. How far did the cable car travel when coming to a stop? (5 points) f. For how much time was the cable car travelling at constant speed? (5 points) g. Extra Credit) Find the total length of cable that passed through the grip from when the gripman first s.) (5 points) acquired the cable to when the cable car came to rest at the second car stop. Show work. (6 points)

Explanation / Answer

a) 9.50 mi/hr = 9.50*88/60 = 13.933 ft/s

b) using: v2 - u2 = 2as

a= v2/ (2s) = 13.9332/(2*50) = 1.941 ft/s2

c) using: v= u+at

t=v/a = 13.933/1.941

t=7.178 seconds

d) again using: v= u+at

0= 13.933 + a*5

a=-2.786 ft/s2

e)  using: v2 - u2 = 2as

s= -13.9332 / (-2*2.786)

s=34.84 ft

f) distance,d travelled at constant speed = 600-(50+34.84)

d=515.16 ft

time = distance/speed = 515.16/13.933

t=36.974 seconds

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