change in relative distance from d1 to d2 where d1>d2
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Two cyclists start from the same place to ride in the same direction.A starts at noon with a speed of 8km/hr and B starts at 2pm with a speed of 10km/hr.At what times A and B will be 5km apart ?
My thought process:
As A starts early at 12 so it will have already covered 16km(8*2).
so S relative=V relative*t or say 16=2*t1 and thus t1=8.
Now we want Srelative to be 5 so
5=2*t2 and t2=2 and a half hour so they will meet at t1-t2 .
Is this correct process ?
arithmetic
asked Dec 10 at 19:35
satyajeet jha
162
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up vote
3
down vote
favorite
Two cyclists start from the same place to ride in the same direction.A starts at noon with a speed of 8km/hr and B starts at 2pm with a speed of 10km/hr.At what times A and B will be 5km apart ?
My thought process:
As A starts early at 12 so it will have already covered 16km(8*2).
so S relative=V relative*t or say 16=2*t1 and thus t1=8.
Now we want Srelative to be 5 so
5=2*t2 and t2=2 and a half hour so they will meet at t1-t2 .
Is this correct process ?
arithmetic
asked Dec 10 at 19:35
satyajeet jha
162
Your method looks perfect! But $8-2.5 = 5.5$ is the number of hours from 2pm that they will be 5 km away, right ? This means they will be 5 km away precicely at $7:30PM$
– rsadhvika
Dec 10 at 19:58
The answer is $t_1pm t_2$ hours ($A$ ahead/behind $B$ by $5$ km) after $2$ pm, since you measured $t$ hours from the departure of $B$.
– Shubham Johri
Dec 10 at 21:40
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Two cyclists start from the same place to ride in the same direction.A starts at noon with a speed of 8km/hr and B starts at 2pm with a speed of 10km/hr.At what times A and B will be 5km apart ?
My thought process:
As A starts early at 12 so it will have already covered 16km(8*2).
so S relative=V relative*t or say 16=2*t1 and thus t1=8.
Now we want Srelative to be 5 so
5=2*t2 and t2=2 and a half hour so they will meet at t1-t2 .
Is this correct process ?
arithmetic
asked Dec 10 at 19:35
satyajeet jha
162
Two cyclists start from the same place to ride in the same direction.A starts at noon with a speed of 8km/hr and B starts at 2pm with a speed of 10km/hr.At what times A and B will be 5km apart ?
My thought process:
As A starts early at 12 so it will have already covered 16km(8*2).
so S relative=V relative*t or say 16=2*t1 and thus t1=8.
Now we want Srelative to be 5 so
5=2*t2 and t2=2 and a half hour so they will meet at t1-t2 .
Is this correct process ?
arithmetic
arithmetic
asked Dec 10 at 19:35
satyajeet jha
162
asked Dec 10 at 19:35
satyajeet jha
162
asked Dec 10 at 19:35
satyajeet jha
162
asked Dec 10 at 19:35
satyajeet jha
162
asked Dec 10 at 19:35
satyajeet jha
162
162
Your method looks perfect! But $8-2.5 = 5.5$ is the number of hours from 2pm that they will be 5 km away, right ? This means they will be 5 km away precicely at $7:30PM$
– rsadhvika
Dec 10 at 19:58
The answer is $t_1pm t_2$ hours ($A$ ahead/behind $B$ by $5$ km) after $2$ pm, since you measured $t$ hours from the departure of $B$.
– Shubham Johri
Dec 10 at 21:40
add a comment |
Your method looks perfect! But $8-2.5 = 5.5$ is the number of hours from 2pm that they will be 5 km away, right ? This means they will be 5 km away precicely at $7:30PM$
– rsadhvika
Dec 10 at 19:58
The answer is $t_1pm t_2$ hours ($A$ ahead/behind $B$ by $5$ km) after $2$ pm, since you measured $t$ hours from the departure of $B$.
– Shubham Johri
Dec 10 at 21:40
Your method looks perfect! But $8-2.5 = 5.5$ is the number of hours from 2pm that they will be 5 km away, right ? This means they will be 5 km away precicely at $7:30PM$
– rsadhvika
Dec 10 at 19:58
Your method looks perfect! But $8-2.5 = 5.5$ is the number of hours from 2pm that they will be 5 km away, right ? This means they will be 5 km away precicely at $7:30PM$
– rsadhvika
Dec 10 at 19:58
The answer is $t_1pm t_2$ hours ($A$ ahead/behind $B$ by $5$ km) after $2$ pm, since you measured $t$ hours from the departure of $B$.
– Shubham Johri
Dec 10 at 21:40
The answer is $t_1pm t_2$ hours ($A$ ahead/behind $B$ by $5$ km) after $2$ pm, since you measured $t$ hours from the departure of $B$.
– Shubham Johri
Dec 10 at 21:40
add a comment |
3 Answers
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up vote
3
down vote
$$x_1(t)=v_1(t-t_1)+x_0=8(t-0)$$
$$x_2(t)=v_2(t-t_2)+x_0=10(t-2)$$
thus the condition
$$|x_2(t)-x_1(t)|=|2t-20|=5$$
gives two solutions
$$t=12,5 text{ or } t=7,5$$
for example,
at $t=7,5$
$$x_1=8(7,5-0)=60 ; km$$
and
$$x_2=10(7,5-2)=55 ; km$$
answered Dec 10 at 19:52
hamam_Abdallah
37.7k21634
add a comment |
up vote
2
down vote
I'll give an alternative, which might look a bit less tedious.
Assume $t=0$ is at 12 noon.
Since $A$ starts at $t=0$ and goes at a speed of 8, we can express the distance traveled by him as
$$a(t) = 8t$$
But $B$ starts with a delay of 2 units; this means his graph shifts to the right by 2 units :
$$b(t) = 10(t-2)$$
Now that equations are setup, you simply have to solve $|b(t)-a(t)|=5$
answered Dec 10 at 19:48
rsadhvika
1,6531228
add a comment |
up vote
2
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By the time $B$ starts, $A$ is ahead by $16$ km. The velocity of $B$ with respect to $A, v_{BA}=v_B-v_A=2$ kmph. Therefore, the distance between $A$ and $B$ changes at $2$ kmph, and after $t$ hours from $B$'s departure, the distance between them is given by $|16-2t|$.
We want $|16-2t|=5implies t=5.5, 10.5$ hours. Since $B$ started at $2$ pm, they are $5$ km apart at $5.5$ and $10.5$ hours after $2$ pm, that is, at $7:30$ pm, $12:30$ am.
answered Dec 10 at 21:39
Shubham Johri
2,578413
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$$x_1(t)=v_1(t-t_1)+x_0=8(t-0)$$
$$x_2(t)=v_2(t-t_2)+x_0=10(t-2)$$
thus the condition
$$|x_2(t)-x_1(t)|=|2t-20|=5$$
gives two solutions
$$t=12,5 text{ or } t=7,5$$
for example,
at $t=7,5$
$$x_1=8(7,5-0)=60 ; km$$
and
$$x_2=10(7,5-2)=55 ; km$$
answered Dec 10 at 19:52
hamam_Abdallah
37.7k21634
add a comment |
up vote
3
down vote
$$x_1(t)=v_1(t-t_1)+x_0=8(t-0)$$
$$x_2(t)=v_2(t-t_2)+x_0=10(t-2)$$
thus the condition
$$|x_2(t)-x_1(t)|=|2t-20|=5$$
gives two solutions
$$t=12,5 text{ or } t=7,5$$
for example,
at $t=7,5$
$$x_1=8(7,5-0)=60 ; km$$
and
$$x_2=10(7,5-2)=55 ; km$$
answered Dec 10 at 19:52
hamam_Abdallah
37.7k21634
add a comment |
up vote
3
down vote
up vote
3
down vote
$$x_1(t)=v_1(t-t_1)+x_0=8(t-0)$$
$$x_2(t)=v_2(t-t_2)+x_0=10(t-2)$$
thus the condition
$$|x_2(t)-x_1(t)|=|2t-20|=5$$
gives two solutions
$$t=12,5 text{ or } t=7,5$$
for example,
at $t=7,5$
$$x_1=8(7,5-0)=60 ; km$$
and
$$x_2=10(7,5-2)=55 ; km$$
answered Dec 10 at 19:52
hamam_Abdallah
37.7k21634
$$x_1(t)=v_1(t-t_1)+x_0=8(t-0)$$
$$x_2(t)=v_2(t-t_2)+x_0=10(t-2)$$
thus the condition
$$|x_2(t)-x_1(t)|=|2t-20|=5$$
gives two solutions
$$t=12,5 text{ or } t=7,5$$
for example,
at $t=7,5$
$$x_1=8(7,5-0)=60 ; km$$
and
$$x_2=10(7,5-2)=55 ; km$$
answered Dec 10 at 19:52
hamam_Abdallah
37.7k21634
answered Dec 10 at 19:52
hamam_Abdallah
37.7k21634
answered Dec 10 at 19:52
hamam_Abdallah
37.7k21634
answered Dec 10 at 19:52
hamam_Abdallah
37.7k21634
37.7k21634
add a comment |
add a comment |
up vote
2
down vote
I'll give an alternative, which might look a bit less tedious.
Assume $t=0$ is at 12 noon.
Since $A$ starts at $t=0$ and goes at a speed of 8, we can express the distance traveled by him as
$$a(t) = 8t$$
But $B$ starts with a delay of 2 units; this means his graph shifts to the right by 2 units :
$$b(t) = 10(t-2)$$
Now that equations are setup, you simply have to solve $|b(t)-a(t)|=5$
answered Dec 10 at 19:48
rsadhvika
1,6531228
add a comment |
up vote
2
down vote
I'll give an alternative, which might look a bit less tedious.
Assume $t=0$ is at 12 noon.
Since $A$ starts at $t=0$ and goes at a speed of 8, we can express the distance traveled by him as
$$a(t) = 8t$$
But $B$ starts with a delay of 2 units; this means his graph shifts to the right by 2 units :
$$b(t) = 10(t-2)$$
Now that equations are setup, you simply have to solve $|b(t)-a(t)|=5$
answered Dec 10 at 19:48
rsadhvika
1,6531228
add a comment |
up vote
2
down vote
up vote
2
down vote
I'll give an alternative, which might look a bit less tedious.
Assume $t=0$ is at 12 noon.
Since $A$ starts at $t=0$ and goes at a speed of 8, we can express the distance traveled by him as
$$a(t) = 8t$$
But $B$ starts with a delay of 2 units; this means his graph shifts to the right by 2 units :
$$b(t) = 10(t-2)$$
Now that equations are setup, you simply have to solve $|b(t)-a(t)|=5$
answered Dec 10 at 19:48
rsadhvika
1,6531228
I'll give an alternative, which might look a bit less tedious.
Assume $t=0$ is at 12 noon.
Since $A$ starts at $t=0$ and goes at a speed of 8, we can express the distance traveled by him as
$$a(t) = 8t$$
But $B$ starts with a delay of 2 units; this means his graph shifts to the right by 2 units :
$$b(t) = 10(t-2)$$
Now that equations are setup, you simply have to solve $|b(t)-a(t)|=5$
answered Dec 10 at 19:48
rsadhvika
1,6531228
answered Dec 10 at 19:48
rsadhvika
1,6531228
answered Dec 10 at 19:48
rsadhvika
1,6531228
answered Dec 10 at 19:48
rsadhvika
1,6531228
1,6531228
add a comment |
add a comment |
up vote
2
down vote
By the time $B$ starts, $A$ is ahead by $16$ km. The velocity of $B$ with respect to $A, v_{BA}=v_B-v_A=2$ kmph. Therefore, the distance between $A$ and $B$ changes at $2$ kmph, and after $t$ hours from $B$'s departure, the distance between them is given by $|16-2t|$.
We want $|16-2t|=5implies t=5.5, 10.5$ hours. Since $B$ started at $2$ pm, they are $5$ km apart at $5.5$ and $10.5$ hours after $2$ pm, that is, at $7:30$ pm, $12:30$ am.
answered Dec 10 at 21:39
Shubham Johri
2,578413
add a comment |
up vote
2
down vote
By the time $B$ starts, $A$ is ahead by $16$ km. The velocity of $B$ with respect to $A, v_{BA}=v_B-v_A=2$ kmph. Therefore, the distance between $A$ and $B$ changes at $2$ kmph, and after $t$ hours from $B$'s departure, the distance between them is given by $|16-2t|$.
We want $|16-2t|=5implies t=5.5, 10.5$ hours. Since $B$ started at $2$ pm, they are $5$ km apart at $5.5$ and $10.5$ hours after $2$ pm, that is, at $7:30$ pm, $12:30$ am.
answered Dec 10 at 21:39
Shubham Johri
2,578413
add a comment |
up vote
2
down vote
up vote
2
down vote
By the time $B$ starts, $A$ is ahead by $16$ km. The velocity of $B$ with respect to $A, v_{BA}=v_B-v_A=2$ kmph. Therefore, the distance between $A$ and $B$ changes at $2$ kmph, and after $t$ hours from $B$'s departure, the distance between them is given by $|16-2t|$.
We want $|16-2t|=5implies t=5.5, 10.5$ hours. Since $B$ started at $2$ pm, they are $5$ km apart at $5.5$ and $10.5$ hours after $2$ pm, that is, at $7:30$ pm, $12:30$ am.
answered Dec 10 at 21:39
Shubham Johri
2,578413
By the time $B$ starts, $A$ is ahead by $16$ km. The velocity of $B$ with respect to $A, v_{BA}=v_B-v_A=2$ kmph. Therefore, the distance between $A$ and $B$ changes at $2$ kmph, and after $t$ hours from $B$'s departure, the distance between them is given by $|16-2t|$.
We want $|16-2t|=5implies t=5.5, 10.5$ hours. Since $B$ started at $2$ pm, they are $5$ km apart at $5.5$ and $10.5$ hours after $2$ pm, that is, at $7:30$ pm, $12:30$ am.
answered Dec 10 at 21:39
Shubham Johri
2,578413
answered Dec 10 at 21:39
Shubham Johri
2,578413
answered Dec 10 at 21:39
Shubham Johri
2,578413
answered Dec 10 at 21:39
Shubham Johri
2,578413
2,578413
add a comment |
add a comment |
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Your method looks perfect! But $8-2.5 = 5.5$ is the number of hours from 2pm that they will be 5 km away, right ? This means they will be 5 km away precicely at $7:30PM$
– rsadhvika
Dec 10 at 19:58
The answer is $t_1pm t_2$ hours ($A$ ahead/behind $B$ by $5$ km) after $2$ pm, since you measured $t$ hours from the departure of $B$.
– Shubham Johri
Dec 10 at 21:40