4) A fish swimming in a horizontal plane has velocity v, (4.0i + 1.0j) m/s at a

Question

4) A fish swimming in a horizontal plane has velocity v, (4.0i + 1.0j) m/s at a point in the ocean whose position vector is r (10i- 4.0) m relative to a stationary rock at the shore. After with constant acceleration for 20.0 s, its velocity is v-(20.0 1-5.0j) m/s. the fish swims (a) What are the components of the acceleration? (b) What is the direction of the acceleration with respect to the fixed +x axis? (c) If the fish maintains constant acceleration, where is it at t- 25.0 s.? (a) A 5.0-kg mass attached to the end of a string swings in a vertical circle of radius 2.5 m. When the ng makes an angle of 350 with the vertical as shown, the speed of the mass is 5.0 m/'s. At this instant t is the magnitude of the force the string exerts on the mass?

Explanation / Answer

a = (v-vi)/t

= [(20.0 i - 5.00 j) - (4.00 i + 1.00 j)]/20

= [16i - 6j]/20
ax = 16.0/20.0 m/s2
ay = -6.0/20.0 m/s2

(b)
tan^-1(-6/16) = -20.56°
-20.56° + 360° = 339.44°

(c) r = ri + (vi)t + (1/2)at^2
= (10.0 i - 4.00 j) + 25(4.00 i + 1.00 j) + (25.0^2)(1/2)[16.0i - 6.0j]/20
= (10.0 i - 4.00 j) + (100.00 i + 25.00 j) + (625)[8.0i - 3.0j]/20
= (110.00 i + 21.00 j) + [250i - 93.75j]
= (360i - 72.75j) m
x = 360 m
y = -72.75 m

v = vi + at = (4.00 i + 1.00 j) + 25[16i - 6j]/20
= (24i - 6.5j) m/s
tan^-1(-6.5/24) = -15.154°
-15.154° + 360° = 344.845°

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