Finding time from a velocity vector

AI Thread Summary
The discussion revolves around calculating time using relative velocities in a scenario involving a boat and water current. The initial confusion stems from an example that used specific numbers not present in the original problem. It clarifies that subtracting velocities is necessary to find the boat's speed relative to the shore, especially when considering currents. The equation provided indicates that time is calculated using the boat's velocity relative to the shore, not the water. The meaning of the distance notation "5.0 km [W20.0S]" remains unclear and is a point of contention.
ericcy
Messages
19
Reaction score
1
Homework Statement
A jet-ski driver wants to head to an island in the St.Lawrence River that is 5.0km [W20.0S] away. If he is traveling at a speed of 40.0km/h relative to the water and the St.Lawrence is flowing 6.0km/h [E], how long will it take him to reach the island?
Relevant Equations
v=d/t
I've looked it up online and someone did t=40−65=0.15(h)

I was just wondering why they would subtract the velocities. Could something explain this to me please? thanks.
 
Physics news on Phys.org
Presumably 5.0km [W20.0S] away means 5.0 km away in the direction 20 degrees West of South. Is this correct?

ericcy said:
I've looked it up online and someone did t=40−65=0.15(h)
That's not very helpful, there are no numbers like 40 or 65 in this problem, so I cannot explain what "they" did and why. The idea of subtracting, or more correctly adding the negative of, velocities is what is done to calculate relative velocities. For example, if your velocity relative to still water is 10 km/h East and the water current is 6 km/h West, your velocity relative to the shore is 10 +(-6) = 4 km/ East. It takes you longer to go the same distance between two fixed points on the shore if you are going against the current and shorter with the current. I am not sure if this is a one or two dimensional relative velocity problem. It depends on what 5.0km [W20.0S] means.
 
kuruman said:
For example, if your velocity relative to still water is 10 km/h East and the water current is 6 km/h West, your velocity relative to the shore is 10 +(-6) = 4 km/ East. It takes you longer to go the same distance between two fixed points on the shore if you are going against the current and shorter with the current. I am not sure if this is a one or two dimensional relative velocity problem. It depends on what 5.0km [W20.0S] means.

If we were to solve for time, like in this question, would we always be expected to use the velocity of the boat relative to the shore and not the water?
 
And my apologies, the equation pasted in wrong, they did t=5[distance to island]/(40[velocity relative to water]-6[velocity of water relative to shore]=0.15h)
 
ericcy said:
If we were to solve for time, like in this question, would we always be expected to use the velocity of the boat relative to the shore and not the water?
Yes because the distance is between points fixed on the shore. If a car drives from point A to point B in such a manner as to keep abreast of the boat, their travel times will be the same, no?

You did not explain what 5.0km [W20.0S] means.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top