A man wishes to travel due North in other to cross a river 5km wide flowing due east at 3km/hr. If he can row as10km/hr in still water find either by scale or calculation.Find:(1)the direction in which he must head his boat in order to get to his destination directly opposite it's starting point (2)the resultant velocity of the boat in the river(3)the time taken to cross the river(4)how far from his destination he would land if he ignorantly stirred North ward
(1) To get to his destination directly opposite his starting point, the man must head his boat in a northeast direction. This way, his rowing speed to the north and the river's speed to the east will cancel each other out.
(2) To calculate the resultant velocity of the boat in the river, we can use the Pythagorean theorem. Let's call the rowing speed of the man in still water "v" km/hr.
The velocity of the boat due north is v km/hr, while the velocity of the river due east is 3 km/hr. Using the Pythagorean theorem, we can find the resultant velocity:
Resultant velocity = sqrt((v^2) + (3^2)) km/hr
(3) To find the time taken to cross the river, we can use the formula:
Time = Distance / Velocity
In this case, the distance is 5 km (width of the river), and the velocity is the resultant velocity calculated in (2). Plug in the values to find the time taken.
(4) If the man ignorantly rowed northward without accounting for the river's flow, he would end up downstream from his destination. The distance he would land from his destination can be calculated by multiplying the time taken to cross the river (found in (3)) by the river's speed:
Distance = Time * River's speed
To solve this problem, we can break it down into different components and use vector addition to find the answers. Let's go step by step:
1) The direction in which he must head his boat to get to the destination directly opposite its starting point:
To get to the destination directly opposite, the man needs to counteract the velocity of the river. Since the river is flowing due east at 3 km/hr, the man needs to row westward with a velocity of 3 km/hr to cancel out the effect of the river's current. Since the man can row at 10 km/hr in still water, he needs to head his boat westward.
2) The resultant velocity of the boat in the river:
To find the resultant velocity, we can use vector addition. The man's rowing velocity in still water is 10 km/hr to the west, and the river's velocity is 3 km/hr to the east. The resultant (net) velocity of the boat will be the difference between the two velocities: 10 km/hr - 3 km/hr = 7 km/hr to the west.
3) The time taken to cross the river:
To find the time taken to cross the river, we need to first calculate the distance to be covered. The width of the river is given as 5 km. Since the boat is moving westward at a velocity of 7 km/hr, the time taken to cross the river would be 5 km / 7 km/hr = 0.714 hours (or approximately 43 minutes).
4) How far from his destination he would land if he ignorantly rowed northward:
If the man ignorantly rows northward, without considering the effect of the river's current, he would be moving perpendicular to the river's current. In this case, the only motion that would contribute to moving across the wider river would be the river's current itself, which is flowing due east at a velocity of 3 km/hr. The time taken to cross the river would then be the same as in step 3 (0.714 hours or approximately 43 minutes). Multiplying the time by the river's velocity will give us the distance from the destination - 0.714 hours * 3 km/hr = 2.142 km.
So, to summarize:
(1) The man must head his boat westward.
(2) The resultant velocity of the boat in the river is 7 km/hr westward.
(3) The time taken to cross the river is approximately 43 minutes.
(4) Ignorantly rowing northward, the man would land approximately 2.142 km from his destination.
To determine the direction, resultant velocity, time taken, and potential landing point, we can use vector addition.
(1) Direction:
To head directly opposite the starting point, the man needs to aim his boat in a direction such that the resultant velocity has no eastward component. This means the boat needs to counteract the eastward component caused by the river.
(2) Resultant Velocity:
The man's speed in still water is 10 km/hr. The river is flowing eastward at 3 km/hr.
Let's denote the boat's velocity in still water as Vb and the river's velocity as Vr.
The eastward component of the resultant velocity is Vr = 3 km/hr, and the northward component is Vb.
Using the Pythagorean theorem, we can find the magnitude of the resultant velocity Vr:
Vr^2 = Vr^2 + Vb^2
3^2 = 3^2 + Vb^2
9 = 9 + Vb^2
Vb^2 = 0
This means the boat needs to head directly north, with no eastward component, in order to counteract the river's current.
(3) Time taken to cross the river:
The river is 5 km wide, and the boat's velocity in still water is 10 km/hr. To calculate the time taken to cross the river, we can use the formula:
Time = Distance / Velocity
Time = 5 km / 10 km/hr
Time = 0.5 hours
(4) Potential landing point if heading north:
If the boat ignores the eastward flow of the river and simply heads north, it will still be affected by the river's current.
In 0.5 hours (the time taken to cross the river), the boat will be carried 0.5 * 3 = 1.5 km eastward by the river's current.
Therefore, the boat will land 1.5 km to the east of its destination if it ignorantly steers northward.
Summary:
(1) The man must head his boat directly north to get to his destination directly opposite.
(2) The resultant velocity of the boat in the river is 10 km/hr directly north.
(3) The time taken to cross the river is 0.5 hours.
(4) If the man ignorantly steers northward, he will land 1.5 km to the east of his destination.