Question

A submarine was floating on the surface on water. It then descended at a rate of 0.5 m/s for 180 s. Then, it ascended at a rate of 0.7 m/s for 75 s. Find the overall and average change in altitude.​

Answer 1

Answer:

Observation:-

• First it is at the surface of water so initial depth is 0

Calculation for the descended:-

Initial height=0m

S=0.5m/s

T=180s

So, directly S=D×T

=0.5×180

= -90 m(depth)

Calculation for the ascend:-

Initial depth= -90m

S=0.7m/

D=S×T

= 75×0.7

= 52.5m

So it gone 52.5 m above

Final height=Initial height-Height gone above

= -90m+52.5m

= -37.5m

So,overall change in altitude= 90+52.5

=142.5m, in which 90m is descend and 52.5 is ascend.

Average change in height:-

$= \frac{90 + 75}{2}$

=165/2

=82.5m