Question

A jogger runs in a straight line with anaverage velocity of 5 m/s for 3.6 min, and then with an average velocity of 3.4 m/s for 2.5 min. What is her total displacement? Answer in units of m.

Answer 1

Data

---------

Velocity one = 5m/s

time one = 3.6min. =216seconds

Velocity two = 3.4m/s

time two = 2.5min. = 150seconds

Solution

-------------

Velocity ( V m/s ) of a particle is the ratio of displacement ( X meters ) to the time ( t seconds )

i.e V = X/t

therefore: X one = (V one)(t one)

= (5m/s)(216s)

=1,080m

X two = (V two)(t two)

= (3.4m/s)(150s)

= 510m

Total displacement

-----------------------------

X total = X one + X two

= 1,080m + 510m

= 1,590m