Question

Question 1 (1 point) Adam and his pet shark spend all day at the beach together. One afternoon, Adam decides to go for a swim. He swims 8.0 x 10+ m [N] in 180 seconds, then swims 0.30 km [W] in 1.5 minutes. [JYVWZR] a. Calculate the total distance he swims. (Be careful with the units.) b. His shark is smart and always swims along the shortest path possible to get back to shore when they go on these trips. Find the magnitude and direction of the shark's path. c. Determine Adam's average speed for the journey. d. Determine the average velocity of Adam's shark.

Answer 1

Adam sulims 4 8.0 x 10 towards north in 180s, then swims 0.30 km towards mest in 1.5 minutes If we draus Adam's trip N BR 10 W ve А A is the point where they start and B is their destination. This makes a right angle at LAOB Hence the shords shortest path to get to B would be the hypotenuse of the triangle A AOB, i.e. BA © a) The total distance that Adam swims is AO + OB = (8 x 10 m m + (0.30 x 10 X 10%), =80m + 3oom = 380 mitre 06) As me discussed earlier the shortest path would be BA B A = OA” JOA² + OB² 180)2 + (300)2 = 16400 +90,000 HA wo = 96,400 =.310.48 mitre Hence, the magnitude of Shark's path is 310.48 m

To find the angle LOAB (0) me have 15 tano = OB OA 300 m 8pm 4 4 or tan (15) = 75.1° .. ZOAB 10 = 75.1° :. The shark swims at an angle of 90° - 75·10) = 14.90 with the horizontal and it is directed towards North-West.

) c) Total distance travelled by Adam is Adam is OA +OB = 320 This distance is travelled by him in time. tott te= = 1808 + 1.5X60) $ = (180 +90) $ 270 seconds Hence, average speed distance travelled norocil time 380 270 Elu = 1.41 m/s det at 03

0 d) The journey The time taken by Adam's shark in the is not given in the question, so by convension that it takes the we take here same time as Adam does.

Hence time taken by shark is to 270 seconds velocity of the shark is ū = ū= displacement time ū 310.48 (m) 1015 m/s along 0= 14.00 270 with the horizontal Velocity is the speed with direction. Answers: a) Total distance Adam swims = 380 m. Magnitude and direction of shark's path is BÀ = 310.48 m at an angle 14.9° mith the horizontal. c) Adam's average speed for the journey is = 415 4/5 V = 1'41 m/s d) Adam's shark's average velocity is ū = 1:15 m/s (14.9° with the horizontal)