A woman is going to a friends house to discuss opening a business. At 11am she s

Question

A woman is going to a friends house to discuss opening a business. At 11am she starts from rest and accelerates at a constant 2.5m/s2 for 9s to get to cruising speed. She drives for 15 minutes at this constant speed before she enters city traffic. She stops at her friends house which is straight lined from hers 27km. She arrives at 12:15. Time interval is 11 to 12:15.

What is her initial velocity?

What is her final velocity ?

What is her average velocity?

What is her velocity 9s into the trip?

What is her average velocity?

What is her velocity 9s into the trip?

Explanation / Answer

suppose initial speed is u m/s.

then after 9s, using v = u + at

v = u + (9*2.5) = u + 22.5

distance traveled in 9 sec = ut + at^2 /2

= 9u + (2.5*9^2 /2) = 9u + 101.25 m

constant speed is u +22.5.

distance traveled in (15min = 900s )

= (u + 22.5)900 = 900u + 20250

total distance traveled = 9u + 101.25 m + 900u + 20250 = 27 x 10^3 m

u = 7.31 m/s

initial velocity = 7.31 m/s or 26.33 km/h ......Ans

final velcoity = 7.31+22.5 = 29.81 m/s or 107.32 km/h ....Ans

average velcity = displacement / time

= (2 x 27)/ ( 1.25 hr) = 43.2 km/h ..Ans

velocity for 9 s trip = (7.31 + 29.81) /2 = 18.56 m/s or 66.82 km/h.....Ans

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