after accelerating down a 26m long hill in 4.0 sa bicyclist is moving 43km/h 5.

Question

after accelerating down a 26m long hill in 4.0 sa bicyclist is moving 43km/h

5. s, Starting from rest, a bus takes 15.0 s to accelerate over 25.0 m. a) What is the final speed (in km/h)? b) What is the average speed (in km/h)? 6. After accelerating down a 26 m long hill in 4.0 s, a bicyclist is moving 43 km/h IN. a) What is the initial velocity? b) What is the average velocity? 7. Before accelerating at 2.78 m/s? over a distance of 111 m, a fire truck was moving 35.2 km/h. a) For how long did the truck accelerate? b) What was the truck's final speed (in km/h)? Answers: 1. a) 66 m b) 11 m/s 2. a) 220 m [Sl b) 87.5 km/h [S 3. a) 4.0s b) 3.5 m/s 4. a) 3.50s b) 87.5 km/h 5. a) 12.0 km/h b) 6.00 km/h 6. a) 1.0 m/s [N] b) 6.5 m/s [N] 7. a) 6.09s b) 96.1 km/h

Explanation / Answer

5. given

starting from rest, the bus takes t = 15 s to accelerate a distance d= 25 m

a) final speed = v

inital speed = u = 0 m/s

then

d = 0.5at^2

25 = 0.5*a*15^2

a = 0.222 m/s/s

hence

2*a*d = v^2

v = 3.333 m/s = 11.988 km/hr

b) average speed vav = d/t = 25/15 = 1.6667 m/s

vav = 6.00012 km/hr

6. d = 26 m, t = 4 s, v = 43 km/h = 11.944 m/s

a. let his initial velocity be u

then

v = u + at

11.944 = u + 4a [ a is acceleration of the man]

ans

2*a*d = v^2 - u^2

52a = 11.944^2 - (11.944 - 4a)^2

52a = 11.944^2 - 11.944^2 - 16a^2 + 95.552a

43.552a = 16a^2

a = 2.722 m/s/s

hence u = v - at = 1.056 m/s = 3.8016 km/hr

b. average velocity = 3.8016 + 0.026*60*60/4 = 27.2016 km/h

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