A certain cable car in San Francisco can stop in 10 s when traveling at maximum speed.?

Let the initial maximum velocity of the car is u m/s,

=>By v = u - at

=>0 = u - a x 10

=>u = 10a ---------(i)

The car comes to the halt after (d+4) meter distance and take 8 sec to travel d meter,

=>By s = ut - 1/2at^2

=>d = 10a x 8 - 1/2 x a x (8)^2

=>d = 80a - 32a

=>d = 48a -----------(iii)

By v^2 = u^2 - 2as

=>0 = (10a)^2 - 2 x a x (d+4)

=>100a^2 = 2ad + 8a

=>100a^2 = 2a x 48a + 8a

=>4a^2 - 8a = 0

=>4a(a -2) = 0

=>a = 0 or 2 m/s^2

=>as a ≠ 0

=>a = 2 m/s^2

By putting this value in (ii) :-

=>d = 48 x 2 = 96 m

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A certain cable car in San Francisco can stop in 10 s when traveling at maximum speed.?

A certain cable car in San Francisco can stop in 10 s when traveling at maximum speed. On one occasion when the cable car is traveling at maximum speed, the driver sees a dog a distance d meters in front of the car and slams on the brakes instantly. The car decelerates constantly and reaches the...

t2 = 10 - 7.44 = 2.56 sec to stop car after passing dog a = 2x/t2² = 2*3.67/2.56² = -1.12 m/s² D = ½a*10² = 56 m (total distance traveled) d = D - 3.67 = 52.33 m

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Let the initial velocity of the car is u m/s =>By v = u - at =>0 = u - 10a =>u = 10a ---------------(i) By s = ut - 1/2at^2 =>d = 10a x 7.44 - 1/2 x a x (7.44)^2 =>d = 46.72a ------------(ii) By v^2 = u^2 - 2as =>0 = (10a)^2 - 2 x a x (46.72a + 3.67) =>100a^2 - 93.44a^2 - 7.34a = 0 =>a(6.56a - 7.34) = 0 =>a = 1.12 m/s^2 Thus d = 46.72 x 1.12 = 52.28 m