1. california sea lions can swim as fast as 40.0 km/h. suppose a sea lion begins to chase a fish at this speed when the fish is 60.0 m away. the fish, of course, does not wait, and swims away at a speed of 16.0 km/h. how long would it take the sea lion to catch the fish? 2. a pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of north. what is the velocity of the dog relative to the road? explain your work and give the formulas you used. tysm < 3!

#1

In order to chase the fish the distance traveled by sea lion in time t must be equal to the distance of sea lion from the fish and distance traveled by fish in the same time.

So here we can say let say sea lion chase the fish in time "t"

then here we have

here

d1 = distance covered by sea lion in time t

d2 = distance covered by fish in the same time t

L = distance between fish and sea lion initially = 60 m

So it will take 9 s to chase the fish by sea lion

# 2

velocity of truck on road = 25 m/s along North

velocity of dog inside the truck = 1.75 m/s at 35 degree East of North

we can write the relative velocity as

now plug in the velocity of truck in this

so it is given as

direction will be given as

so with respect to ground dog velocity is 26.44 m/s towards 2.2 degree East of North

spectral lines lyman is one set