(Q4 – Two-Dimensional Motion with Constant Acceleration) A fish swimming in a ho
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Question
(Q4 – Two-Dimensional Motion with Constant Acceleration)
A fish swimming in a horizontal plane has velocity Vo= 4.00i+1. 00j m/s at a point in the ocean where the position relative to a certain rock is Ro= 10.00i-4. 00j m. After the fish swims with constant acceleration or 20.0 s. its velocity is Vf= 20.00i-5. 00j m/s.
Show your work: (a) What are the components of the acceleration? (b) If the fish maintains its constant acceleration determined in part (b), where is it at t=25.0 s. (c) What is the velocity of the fish at t = 25.0 s (d) What is the direction (measured counter clockwise from x-axis) of the fish movement at t = 25.0 s.
Explanation / Answer
vo = 4.0 i +1.0 j m/s
ro = 10.0 i - 4.0 j m
t = 20.0 s
vf = 20.0 i - 5.0 j m/s
vf = vo + a*t
In X direction,
20.0 = 4.0 + ax * 20.0
ax = 0.8 i^ m/s^2
In Y direction,
- 5.0 = 1.0 + ay*20.0
ay = - 0.3 j^ m/s^2
a = 0.8 i^ - 0.3 j^ m/s^2
(b)
At t = 25.0 s,
r = ro + vo*t + 1/2 * a * t^2
r = 10.0 i - 4.0 j + (4.0 i +1.0 j) * 25.0 + 1/2 * (0.8 i^ - 0.3j^) * 25.0^2
r = (10.0 + 4.0 * 25.0 + 1/2*0.8 * 25.0^2) i^ + ( -4.0 + 1.0*25.0 - 1/2*0.3*25.0^2) j^
r = 360 i^ - 72.75 j^ m
(c)
vf = vo + a * t
vf = (4.0 i +1.0 j) + (0.8 i^ - 0.3j^) * 25.0
vf = 24.0 i^ - 6.5 j^ m/s
(d)
Direction = tan^-1(6.5/24.0)
Direction = 345o counter clockwise from x-axis
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