Question

A jogger ran for 15 minutes at 4 mph in the direction 50 degrees South of West. Then she stopped, turned to the East, and ran with a constant acceleration of 0.2 m/s^2 for 1.0 minutes. She stopped again, turned, and ran back to the gym with a constant velocity. In all, she spent 40 minutes running and jogging.

a) What was her velocity (Magnitude and Direction) as she was running back to the gym?

b)Write the vector of the first leg of her trip as: Magnitude at direction = ?

c) Write the vector of the second leg of her trip as: Magnitude at direction = ?

d) Write the vector of her total displacement in the first two legs of her trip as: Magnitude at direction = ?

e) Write the vector of her velocity as she is running back: Magnitude @ Direction = ?

Answer 1

Return Time taken to 40-16=24 mins=24*60=1440 second

One mile = 1.609km

Distance travelled in first 15 min=speed * time

S1= (4×1.609×1000/3600)×(15×60)=1609m south-West direction

Distance covered in East direction=s2=0.5at^2

=0.5×0.2×60^2=360m

S3=return distance =sqrt((1609cos40)^2 +(1609sin40-360)^2) =1404.92m

V3=s3/return time=1404.92/1440=0.975sec

S3 vector = - ((1609sin40-360)i+1609cos40j)=

-674.24i-1232.56j m

Displacement vector before return, s3

Velocity return =s3/return Time m/s

Velocity of first leg=-4sin40i-4cos40j mph

Velocity of second leg=at i=0.2*60 i=12 i

Note: i = +ve x direction