Physics with calc 1 - Answer the (no response) 1.A fish swimming in a horizontal
ID: 1434127 • Letter: P
Question
Physics with calc 1 - Answer the (no response)
1.A fish swimming in a horizontal plane has velocity vi= (4.00 i + 5.00 j) m/s at a point in the ocean where the position relative to a certain rock is ri= (-10.0 i + 4.00 j) m. After the fish swims with constant acceleration for 25.0 s, its velocity is v = (19.0 i - 5.00 j) m/s.
(a) What are the components of the acceleration? What is the direction of the acceleration with respect to unit vector i?
ax = (No Response)
ay = (No Response)
=(No Response) ° (counterclockwise from the +x-axis is positive)
(b) If the fish maintains constant acceleration, where is it at t = 35.0 s, with respect to the rock?
x = (No Response)
y = (No Response)
(c) In what direction is it moving? (Hint: This is NOT asking for the direction of the vector you found in part b. You need to find the velocity vector.)
(No Response) ° (counterclockwise from the +x-axis is positive)
Explanation / Answer
a) a(t) = dV/dt
a(t) = (19i - 5j) - (4i + 5j) / 25
a(t) = 0.6 i - 0.4j
a(x) = 0.6
a(y) = -0.4
= 90 + tan^-1(0.4/0.6) = 146.31 degree from the x-axis in counterclockwise direction
b) S= ut + 0.5at^2
x = 4*35 + 0.5*0.6*35^2 = 507.5 m
y = 5*35 + 0.5*(-0.4)*35^2
y = -70 m
r(t) = 507.5 i - 70 j