The speed of a boat is 5Km/h in still water. It crosses a river of width 1km along the shortest path in 15 minutes.A motorboat going downstream overcame a raft at a point A (Kinematics question)How to find the power of a jet engine?The vertical velocity of a ball hit by a bat is given in the following options. Which of them is right?Equation of motion of a body is $fracdvdt=-4v+8$ where is velocity in $m/s$ and $t$ is time.A particle is fired with velocity u making an angle $theta$ with the horizontal.A particle moves in such a way that its position varies with time t as $r=ti+fract^22j+tk$A person moving east finds wind blowing south. If he moves with thrice the velocity, he finds wind blowing southwest. What's wind speed?
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The speed of a boat is 5Km/h in still water. It crosses a river of width 1km along the shortest path in 15 minutes.
A motorboat going downstream overcame a raft at a point A (Kinematics question)How to find the power of a jet engine?The vertical velocity of a ball hit by a bat is given in the following options. Which of them is right?Equation of motion of a body is $fracdvdt=-4v+8$ where is velocity in $m/s$ and $t$ is time.A particle is fired with velocity u making an angle $theta$ with the horizontal.A particle moves in such a way that its position varies with time t as $r=ti+fract^22j+tk$A person moving east finds wind blowing south. If he moves with thrice the velocity, he finds wind blowing southwest. What's wind speed?
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$begingroup$
Velocity of the river is?
Since it covers 1km in 15 mins, the relative velocity of the boat with respect to the river will be
$$frac 14 V_br=1$$
$$V_br=4km/h$$
So $$V_b=V_br + V_r$$
$$V_r=1km/h$$
The right answer is 3km/h, so what am I doing wrong?
kinematics
asked 8 hours ago
Aditya Aditya
45310 bronze badges
$endgroup$
add a comment |
$begingroup$
Velocity of the river is?
Since it covers 1km in 15 mins, the relative velocity of the boat with respect to the river will be
$$frac 14 V_br=1$$
$$V_br=4km/h$$
So $$V_b=V_br + V_r$$
$$V_r=1km/h$$
The right answer is 3km/h, so what am I doing wrong?
kinematics
asked 8 hours ago
Aditya Aditya
45310 bronze badges
$endgroup$
1
$begingroup$
The flowing water will give lateral movement to the boat. Think of Pythagoras. Your initial calculation $V_br$ will reveal the velocity of water. The equality $V_b = V_br + V_r$ is equality of vectors.
$endgroup$
– Alvin Lepik
8 hours ago
1
$begingroup$
You need to take into account, that velocities are vectors! In this case, it is sufficient to take them as 2-dimensional vectors. The Velocity is the absolute value of the velocity vector. Therefore: $v_b = sqrtv_br ^2 + v_r ^2$. Given $v_br=4$ and $v_b = 5$, you find $v_r=3$. (recall: $3^2+4^2=5^2$). Try to work out the details of the addition (what are the directions of the velocities!)
$endgroup$
– Caroline
8 hours ago
$begingroup$
Yeah, thought that, but since the angles weren’t given, I wasn’t sure about it.
$endgroup$
– Aditya
8 hours ago
add a comment |
$begingroup$
Velocity of the river is?
Since it covers 1km in 15 mins, the relative velocity of the boat with respect to the river will be
$$frac 14 V_br=1$$
$$V_br=4km/h$$
So $$V_b=V_br + V_r$$
$$V_r=1km/h$$
The right answer is 3km/h, so what am I doing wrong?
kinematics
asked 8 hours ago
Aditya Aditya
45310 bronze badges
$endgroup$
Velocity of the river is?
Since it covers 1km in 15 mins, the relative velocity of the boat with respect to the river will be
$$frac 14 V_br=1$$
$$V_br=4km/h$$
So $$V_b=V_br + V_r$$
$$V_r=1km/h$$
The right answer is 3km/h, so what am I doing wrong?
kinematics
kinematics
asked 8 hours ago
Aditya Aditya
45310 bronze badges
asked 8 hours ago
Aditya Aditya
45310 bronze badges
asked 8 hours ago
Aditya Aditya
45310 bronze badges
asked 8 hours ago
Aditya Aditya
45310 bronze badges
asked 8 hours ago
Aditya Aditya
45310 bronze badges
45310 bronze badges
1
$begingroup$
The flowing water will give lateral movement to the boat. Think of Pythagoras. Your initial calculation $V_br$ will reveal the velocity of water. The equality $V_b = V_br + V_r$ is equality of vectors.
$endgroup$
– Alvin Lepik
8 hours ago
1
$begingroup$
You need to take into account, that velocities are vectors! In this case, it is sufficient to take them as 2-dimensional vectors. The Velocity is the absolute value of the velocity vector. Therefore: $v_b = sqrtv_br ^2 + v_r ^2$. Given $v_br=4$ and $v_b = 5$, you find $v_r=3$. (recall: $3^2+4^2=5^2$). Try to work out the details of the addition (what are the directions of the velocities!)
$endgroup$
– Caroline
8 hours ago
$begingroup$
Yeah, thought that, but since the angles weren’t given, I wasn’t sure about it.
$endgroup$
– Aditya
8 hours ago
add a comment |
1
$begingroup$
The flowing water will give lateral movement to the boat. Think of Pythagoras. Your initial calculation $V_br$ will reveal the velocity of water. The equality $V_b = V_br + V_r$ is equality of vectors.
$endgroup$
– Alvin Lepik
8 hours ago
1
$begingroup$
You need to take into account, that velocities are vectors! In this case, it is sufficient to take them as 2-dimensional vectors. The Velocity is the absolute value of the velocity vector. Therefore: $v_b = sqrtv_br ^2 + v_r ^2$. Given $v_br=4$ and $v_b = 5$, you find $v_r=3$. (recall: $3^2+4^2=5^2$). Try to work out the details of the addition (what are the directions of the velocities!)
$endgroup$
– Caroline
8 hours ago
$begingroup$
Yeah, thought that, but since the angles weren’t given, I wasn’t sure about it.
$endgroup$
– Aditya
8 hours ago
1
1
$begingroup$
The flowing water will give lateral movement to the boat. Think of Pythagoras. Your initial calculation $V_br$ will reveal the velocity of water. The equality $V_b = V_br + V_r$ is equality of vectors.
$endgroup$
– Alvin Lepik
8 hours ago
$begingroup$
The flowing water will give lateral movement to the boat. Think of Pythagoras. Your initial calculation $V_br$ will reveal the velocity of water. The equality $V_b = V_br + V_r$ is equality of vectors.
$endgroup$
– Alvin Lepik
8 hours ago
1
1
$begingroup$
You need to take into account, that velocities are vectors! In this case, it is sufficient to take them as 2-dimensional vectors. The Velocity is the absolute value of the velocity vector. Therefore: $v_b = sqrtv_br ^2 + v_r ^2$. Given $v_br=4$ and $v_b = 5$, you find $v_r=3$. (recall: $3^2+4^2=5^2$). Try to work out the details of the addition (what are the directions of the velocities!)
$endgroup$
– Caroline
8 hours ago
$begingroup$
You need to take into account, that velocities are vectors! In this case, it is sufficient to take them as 2-dimensional vectors. The Velocity is the absolute value of the velocity vector. Therefore: $v_b = sqrtv_br ^2 + v_r ^2$. Given $v_br=4$ and $v_b = 5$, you find $v_r=3$. (recall: $3^2+4^2=5^2$). Try to work out the details of the addition (what are the directions of the velocities!)
$endgroup$
– Caroline
8 hours ago
$begingroup$
Yeah, thought that, but since the angles weren’t given, I wasn’t sure about it.
$endgroup$
– Aditya
8 hours ago
$begingroup$
Yeah, thought that, but since the angles weren’t given, I wasn’t sure about it.
$endgroup$
– Aditya
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You should use the equation
$$V_b^2=V_br^2+ V_r^2$$
instead of
$$V_b=V_br + V_r$$
because the three velocity vectors are not in the same direction. Rather, they form a right triangle.
Then, you get
$$V_r = sqrt V_b^2 - V_br^2 = sqrt 5^2 - 4^2= 3 textkm/h$$
answered 7 hours ago
QuantoQuanto
3,7351 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
The only way to go along the shortest path is by tilting the boat against the river current. A good picture helps:
You're given $V_BW = 5kmph$ and $V_BS = 1km ~in~15min$ .
You can find $V_WS$
answered 8 hours ago
ganeshie8ganeshie8
5,5101 gold badge11 silver badges19 bronze badges
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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votes
active
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active
oldest
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$begingroup$
You should use the equation
$$V_b^2=V_br^2+ V_r^2$$
instead of
$$V_b=V_br + V_r$$
because the three velocity vectors are not in the same direction. Rather, they form a right triangle.
Then, you get
$$V_r = sqrt V_b^2 - V_br^2 = sqrt 5^2 - 4^2= 3 textkm/h$$
answered 7 hours ago
QuantoQuanto
3,7351 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
You should use the equation
$$V_b^2=V_br^2+ V_r^2$$
instead of
$$V_b=V_br + V_r$$
because the three velocity vectors are not in the same direction. Rather, they form a right triangle.
Then, you get
$$V_r = sqrt V_b^2 - V_br^2 = sqrt 5^2 - 4^2= 3 textkm/h$$
answered 7 hours ago
QuantoQuanto
3,7351 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
You should use the equation
$$V_b^2=V_br^2+ V_r^2$$
instead of
$$V_b=V_br + V_r$$
because the three velocity vectors are not in the same direction. Rather, they form a right triangle.
Then, you get
$$V_r = sqrt V_b^2 - V_br^2 = sqrt 5^2 - 4^2= 3 textkm/h$$
answered 7 hours ago
QuantoQuanto
3,7351 silver badge12 bronze badges
$endgroup$
You should use the equation
$$V_b^2=V_br^2+ V_r^2$$
instead of
$$V_b=V_br + V_r$$
because the three velocity vectors are not in the same direction. Rather, they form a right triangle.
Then, you get
$$V_r = sqrt V_b^2 - V_br^2 = sqrt 5^2 - 4^2= 3 textkm/h$$
answered 7 hours ago
QuantoQuanto
3,7351 silver badge12 bronze badges
edited 4 hours ago
answered 7 hours ago
QuantoQuanto
3,7351 silver badge12 bronze badges
answered 7 hours ago
QuantoQuanto
3,7351 silver badge12 bronze badges
answered 7 hours ago
QuantoQuanto
3,7351 silver badge12 bronze badges
3,7351 silver badge12 bronze badges
add a comment |
add a comment |
$begingroup$
The only way to go along the shortest path is by tilting the boat against the river current. A good picture helps:
You're given $V_BW = 5kmph$ and $V_BS = 1km ~in~15min$ .
You can find $V_WS$
answered 8 hours ago
ganeshie8ganeshie8
5,5101 gold badge11 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
The only way to go along the shortest path is by tilting the boat against the river current. A good picture helps:
You're given $V_BW = 5kmph$ and $V_BS = 1km ~in~15min$ .
You can find $V_WS$
answered 8 hours ago
ganeshie8ganeshie8
5,5101 gold badge11 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
The only way to go along the shortest path is by tilting the boat against the river current. A good picture helps:
You're given $V_BW = 5kmph$ and $V_BS = 1km ~in~15min$ .
You can find $V_WS$
answered 8 hours ago
ganeshie8ganeshie8
5,5101 gold badge11 silver badges19 bronze badges
$endgroup$
The only way to go along the shortest path is by tilting the boat against the river current. A good picture helps:
You're given $V_BW = 5kmph$ and $V_BS = 1km ~in~15min$ .
You can find $V_WS$
answered 8 hours ago
ganeshie8ganeshie8
5,5101 gold badge11 silver badges19 bronze badges
answered 8 hours ago
ganeshie8ganeshie8
5,5101 gold badge11 silver badges19 bronze badges
answered 8 hours ago
ganeshie8ganeshie8
5,5101 gold badge11 silver badges19 bronze badges
answered 8 hours ago
ganeshie8ganeshie8
5,5101 gold badge11 silver badges19 bronze badges
5,5101 gold badge11 silver badges19 bronze badges
add a comment |
add a comment |
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$begingroup$
The flowing water will give lateral movement to the boat. Think of Pythagoras. Your initial calculation $V_br$ will reveal the velocity of water. The equality $V_b = V_br + V_r$ is equality of vectors.
$endgroup$
– Alvin Lepik
8 hours ago
1
$begingroup$
You need to take into account, that velocities are vectors! In this case, it is sufficient to take them as 2-dimensional vectors. The Velocity is the absolute value of the velocity vector. Therefore: $v_b = sqrtv_br ^2 + v_r ^2$. Given $v_br=4$ and $v_b = 5$, you find $v_r=3$. (recall: $3^2+4^2=5^2$). Try to work out the details of the addition (what are the directions of the velocities!)
$endgroup$
– Caroline
8 hours ago
$begingroup$
Yeah, thought that, but since the angles weren’t given, I wasn’t sure about it.
$endgroup$
– Aditya
8 hours ago