Question

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?

Please explain your work and give the formulas you used. Tysm <3!

Answer 1

#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

$d_1 = d_2 + L$

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

$d_1 = v_1 * t$

$d_1 = (40*\frac{5}{18})*t = \frac{100}{9}*t$

$d_2 = (16*\frac{5}{18})*t = \frac{40}{9}*t$

$\frac{100}{9}*t = \frac{40}{9}*t + 60$

$\frac{100}{9}*t - \frac{40}{9}*t = 60$

$\frac{60}{9}*t = 60$

$t = 9 s$

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

$v_{dt} = 1.75 cos35\hat j + 1.75sin35 \hat i$

$v_{dt} = 1.43 \hat j + 1 \hat i$

we can write the relative velocity as

$v_d - v_t = 1.43 \hat j + 1 \hat i$

$v_d = v_t + (1.43 \hat j + 1 \hat i)$

now plug in the velocity of truck in this

$v_d = 25 \hat j + (1.43 \hat j + 1 \hat i)$

$v_d = 26.43 \hat j + 1 \hat i$

so it is given as

$v_d = \sqrt{26.43^2 + 1^2} = 26.44 m/s$

direction will be given as

$\theta = tan^{-1}\frac{v_x}{v_y}$

$\theta = tan^{-1}\frac{1}{26.43} = 2.2 degree$

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

Answer 2

Answer:

1) time t = 9 s

2) velocity v  = 26.4 m/s 2.17° east of north

Explanation:

1

Given

velocity of sea lion $v_s = 40.0 kmph$

velocity of the fish $v-f = 16 kmph$

Distance  between the fish and sea lion d = 60 m

Solution

Relative velocity of the sea lion with respect to the fish

$v = v_s - v_f\\\\v = 40 - 16\\\\v = 24.kmph\\\\v = 24 \times \frac{1000}{60 \times 60} m/s\\\\v = 6.67 m/s$

Time

$t = \frac{d}{v} \\\\t = \frac{60}{6.67} \\\\t = 9 s$

2

Given

velocity of the truck $v_t = 25 m/s$

velocity of the dog $v_d = 1.75 m/s$

Angle $\theta = 35.0^o$

Solution

Dog moves in north east so

Component of velocity in East

$v_{d,E} = v_dsin\theta\\\\v_{d,E} = 1.75 \times sin35\\\\v_{d,E} = 1.004 m/s$

Component of velocity in North

$v_{d,N} = v_dcos\theta\\\\v_{d,N} = 1.75 \times cos35\\\\v_{d,N} = 1.434 m/s$

The velocity of Dog relative to the road = velocity of the dog relative to truck + velocity of the truck relative to the road

Y component

$v_y = v_t + v_{d,N}\\\\v_y = 25 + 1.434\\\\v_y = 26.434 m/s$

X component

$v_x = v_{d,E}\\\\V_x = 1.004 m/s$

Velocity

$v = \sqrt{v_x^{2} +v_y^{2} } \\\\v = \sqrt{1.004^2 + 26.434^2} \\\\v = 26.4 m/s$

Direction

$\phi = tan^{-1}\frac{v_x}{v_y} \\\\\phi = tan^{-1}\frac{1.004}{26.434} \\\\\phi = 2.17^o east of north$