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?

Answer 1

Answer:

a) Since the woman starts from rest hence her initial velocity equals 0 m/s.

b) We can calculate the final velocity as follows

1) Distance covered while accelerating for 9 seconds is

$s_1=\frac{1}{2}\times at^{2}\\\\s=0.5\times 2.5\times 9^{2}=101.25m$

2) Cruising speed =$v_{c}=at=v_{c}=2.5\times 9=22.5m/s$

Distance covered while travelling for 15 minutes at cruising speed =

$s_{2}=22.5\times 15\times 60\\\\s_{2}=20.250km$

Since the woman cover's 27 km in total hence the speed she travles with traffic speed can be calculated as

$v_{traffic}=\frac{Distance}{Time}\\\\v_{traffic}=\frac{27000-101.25-20250}{4500-9-900}\\\\v_{traffic}=1.85m/s$

hence the final speed of woman = $1.85 m/s$

Average velocity = $V_{avg}=\frac{Distance}{Time}\\\\V_{avg}=\frac{27000}{4500}m/s\\\\V_{avg}=6m/s$

Her velocity 9 seconds into trip is calculated earlier as 22.5 m/s