Distance covered by two object from different initial point
$begingroup$
Two cars, an Edsel and a Studebaker, are 635 kilometers apart, with Edsel moving behind Studebaker(otherwise they won't never meet, by the speeds given). They start at the same time and drive in one direction . The Edsel travels at a rate of $70$ kilometers per hour and the Studebaker travels $57$
kilometers per hour. In how many hours will the two cars meet?
algebra-precalculus
edited Dec 23 '18 at 11:47
Ankit Kumar
1,494221
asked Dec 2 '18 at 11:42
tomtomtomtom
63
$endgroup$
add a comment |
$begingroup$
Two cars, an Edsel and a Studebaker, are 635 kilometers apart, with Edsel moving behind Studebaker(otherwise they won't never meet, by the speeds given). They start at the same time and drive in one direction . The Edsel travels at a rate of $70$ kilometers per hour and the Studebaker travels $57$
kilometers per hour. In how many hours will the two cars meet?
algebra-precalculus
edited Dec 23 '18 at 11:47
Ankit Kumar
1,494221
asked Dec 2 '18 at 11:42
tomtomtomtom
63
$endgroup$
1
$begingroup$
how many km are removed from the remaining distance every hour ? then apply cross-muiltiplication.
$endgroup$
– zwim
Dec 2 '18 at 11:47
$begingroup$
Divide the total distance traveled by the total distance the two cars travel in one hour.
$endgroup$
– N. F. Taussig
Dec 2 '18 at 11:55
$begingroup$
Depending upon what direction they are traveling they may not meet but in other direction they will meet. At that time what will the impact be?
$endgroup$
– William Elliot
Dec 3 '18 at 0:30
add a comment |
$begingroup$
Two cars, an Edsel and a Studebaker, are 635 kilometers apart, with Edsel moving behind Studebaker(otherwise they won't never meet, by the speeds given). They start at the same time and drive in one direction . The Edsel travels at a rate of $70$ kilometers per hour and the Studebaker travels $57$
kilometers per hour. In how many hours will the two cars meet?
algebra-precalculus
edited Dec 23 '18 at 11:47
Ankit Kumar
1,494221
asked Dec 2 '18 at 11:42
tomtomtomtom
63
$endgroup$
Two cars, an Edsel and a Studebaker, are 635 kilometers apart, with Edsel moving behind Studebaker(otherwise they won't never meet, by the speeds given). They start at the same time and drive in one direction . The Edsel travels at a rate of $70$ kilometers per hour and the Studebaker travels $57$
kilometers per hour. In how many hours will the two cars meet?
algebra-precalculus
algebra-precalculus
edited Dec 23 '18 at 11:47
Ankit Kumar
1,494221
asked Dec 2 '18 at 11:42
tomtomtomtom
63
edited Dec 23 '18 at 11:47
Ankit Kumar
1,494221
asked Dec 2 '18 at 11:42
tomtomtomtom
63
edited Dec 23 '18 at 11:47
Ankit Kumar
1,494221
edited Dec 23 '18 at 11:47
Ankit Kumar
1,494221
edited Dec 23 '18 at 11:47
Ankit Kumar
1,494221
1,494221
asked Dec 2 '18 at 11:42
tomtomtomtom
63
asked Dec 2 '18 at 11:42
tomtomtomtom
63
asked Dec 2 '18 at 11:42
tomtomtomtom
63
63
1
$begingroup$
how many km are removed from the remaining distance every hour ? then apply cross-muiltiplication.
$endgroup$
– zwim
Dec 2 '18 at 11:47
$begingroup$
Divide the total distance traveled by the total distance the two cars travel in one hour.
$endgroup$
– N. F. Taussig
Dec 2 '18 at 11:55
$begingroup$
Depending upon what direction they are traveling they may not meet but in other direction they will meet. At that time what will the impact be?
$endgroup$
– William Elliot
Dec 3 '18 at 0:30
add a comment |
1
$begingroup$
how many km are removed from the remaining distance every hour ? then apply cross-muiltiplication.
$endgroup$
– zwim
Dec 2 '18 at 11:47
$begingroup$
Divide the total distance traveled by the total distance the two cars travel in one hour.
$endgroup$
– N. F. Taussig
Dec 2 '18 at 11:55
$begingroup$
Depending upon what direction they are traveling they may not meet but in other direction they will meet. At that time what will the impact be?
$endgroup$
– William Elliot
Dec 3 '18 at 0:30
1
1
$begingroup$
how many km are removed from the remaining distance every hour ? then apply cross-muiltiplication.
$endgroup$
– zwim
Dec 2 '18 at 11:47
$begingroup$
how many km are removed from the remaining distance every hour ? then apply cross-muiltiplication.
$endgroup$
– zwim
Dec 2 '18 at 11:47
$begingroup$
Divide the total distance traveled by the total distance the two cars travel in one hour.
$endgroup$
– N. F. Taussig
Dec 2 '18 at 11:55
$begingroup$
Divide the total distance traveled by the total distance the two cars travel in one hour.
$endgroup$
– N. F. Taussig
Dec 2 '18 at 11:55
$begingroup$
Depending upon what direction they are traveling they may not meet but in other direction they will meet. At that time what will the impact be?
$endgroup$
– William Elliot
Dec 3 '18 at 0:30
$begingroup$
Depending upon what direction they are traveling they may not meet but in other direction they will meet. At that time what will the impact be?
$endgroup$
– William Elliot
Dec 3 '18 at 0:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Edsel will have to cover $635$ kilometers faster compared to Studebaker in order to meet him. Further, he's moving $70-57=13$ km/h faster than Studebaker. So, he'll meet Studebaker in:
$$T=frac{635text{km}}{13text{km/hr}}=boxed{48.846text{ hr}}$$
answered Dec 23 '18 at 11:30
Ankit KumarAnkit Kumar
1,494221
$endgroup$
add a comment |
$begingroup$
Method 1
Use concept of relative velocity. In the reference frame of Studebaker, velocity of Edsel is $70-57=13kmph$ towards Studebaker. And the relative displacement at the time of meeting is $ 635$ km. Therefore the time taken will be $frac{635}{13}=48.846 hr$, which is the required answer.
Method 2
Let the required answer be $t$ hr.Therefore, at the time of meeting, Studebaker has covered $57t$ km and that covered by Edsen is $70t$km. But we know that the extra distance covered by Essen is $635$ km. Hence,
$$70t-57t=635$$
$$Longrightarrow t=frac{635}{13}=48.846 hr$$
which is the required answer.
Hope it is helpful:)
answered Dec 23 '18 at 11:51
MartundMartund
1,633213
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Edsel will have to cover $635$ kilometers faster compared to Studebaker in order to meet him. Further, he's moving $70-57=13$ km/h faster than Studebaker. So, he'll meet Studebaker in:
$$T=frac{635text{km}}{13text{km/hr}}=boxed{48.846text{ hr}}$$
answered Dec 23 '18 at 11:30
Ankit KumarAnkit Kumar
1,494221
$endgroup$
add a comment |
$begingroup$
Edsel will have to cover $635$ kilometers faster compared to Studebaker in order to meet him. Further, he's moving $70-57=13$ km/h faster than Studebaker. So, he'll meet Studebaker in:
$$T=frac{635text{km}}{13text{km/hr}}=boxed{48.846text{ hr}}$$
answered Dec 23 '18 at 11:30
Ankit KumarAnkit Kumar
1,494221
$endgroup$
add a comment |
$begingroup$
Edsel will have to cover $635$ kilometers faster compared to Studebaker in order to meet him. Further, he's moving $70-57=13$ km/h faster than Studebaker. So, he'll meet Studebaker in:
$$T=frac{635text{km}}{13text{km/hr}}=boxed{48.846text{ hr}}$$
answered Dec 23 '18 at 11:30
Ankit KumarAnkit Kumar
1,494221
$endgroup$
Edsel will have to cover $635$ kilometers faster compared to Studebaker in order to meet him. Further, he's moving $70-57=13$ km/h faster than Studebaker. So, he'll meet Studebaker in:
$$T=frac{635text{km}}{13text{km/hr}}=boxed{48.846text{ hr}}$$
answered Dec 23 '18 at 11:30
Ankit KumarAnkit Kumar
1,494221
answered Dec 23 '18 at 11:30
Ankit KumarAnkit Kumar
1,494221
answered Dec 23 '18 at 11:30
Ankit KumarAnkit Kumar
1,494221
answered Dec 23 '18 at 11:30
Ankit KumarAnkit Kumar
1,494221
1,494221
add a comment |
add a comment |
$begingroup$
Method 1
Use concept of relative velocity. In the reference frame of Studebaker, velocity of Edsel is $70-57=13kmph$ towards Studebaker. And the relative displacement at the time of meeting is $ 635$ km. Therefore the time taken will be $frac{635}{13}=48.846 hr$, which is the required answer.
Method 2
Let the required answer be $t$ hr.Therefore, at the time of meeting, Studebaker has covered $57t$ km and that covered by Edsen is $70t$km. But we know that the extra distance covered by Essen is $635$ km. Hence,
$$70t-57t=635$$
$$Longrightarrow t=frac{635}{13}=48.846 hr$$
which is the required answer.
Hope it is helpful:)
answered Dec 23 '18 at 11:51
MartundMartund
1,633213
$endgroup$
add a comment |
$begingroup$
Method 1
Use concept of relative velocity. In the reference frame of Studebaker, velocity of Edsel is $70-57=13kmph$ towards Studebaker. And the relative displacement at the time of meeting is $ 635$ km. Therefore the time taken will be $frac{635}{13}=48.846 hr$, which is the required answer.
Method 2
Let the required answer be $t$ hr.Therefore, at the time of meeting, Studebaker has covered $57t$ km and that covered by Edsen is $70t$km. But we know that the extra distance covered by Essen is $635$ km. Hence,
$$70t-57t=635$$
$$Longrightarrow t=frac{635}{13}=48.846 hr$$
which is the required answer.
Hope it is helpful:)
answered Dec 23 '18 at 11:51
MartundMartund
1,633213
$endgroup$
add a comment |
$begingroup$
Method 1
Use concept of relative velocity. In the reference frame of Studebaker, velocity of Edsel is $70-57=13kmph$ towards Studebaker. And the relative displacement at the time of meeting is $ 635$ km. Therefore the time taken will be $frac{635}{13}=48.846 hr$, which is the required answer.
Method 2
Let the required answer be $t$ hr.Therefore, at the time of meeting, Studebaker has covered $57t$ km and that covered by Edsen is $70t$km. But we know that the extra distance covered by Essen is $635$ km. Hence,
$$70t-57t=635$$
$$Longrightarrow t=frac{635}{13}=48.846 hr$$
which is the required answer.
Hope it is helpful:)
answered Dec 23 '18 at 11:51
MartundMartund
1,633213
$endgroup$
Method 1
Use concept of relative velocity. In the reference frame of Studebaker, velocity of Edsel is $70-57=13kmph$ towards Studebaker. And the relative displacement at the time of meeting is $ 635$ km. Therefore the time taken will be $frac{635}{13}=48.846 hr$, which is the required answer.
Method 2
Let the required answer be $t$ hr.Therefore, at the time of meeting, Studebaker has covered $57t$ km and that covered by Edsen is $70t$km. But we know that the extra distance covered by Essen is $635$ km. Hence,
$$70t-57t=635$$
$$Longrightarrow t=frac{635}{13}=48.846 hr$$
which is the required answer.
Hope it is helpful:)
answered Dec 23 '18 at 11:51
MartundMartund
1,633213
answered Dec 23 '18 at 11:51
MartundMartund
1,633213
answered Dec 23 '18 at 11:51
MartundMartund
1,633213
answered Dec 23 '18 at 11:51
MartundMartund
1,633213
1,633213
add a comment |
add a comment |
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1
$begingroup$
how many km are removed from the remaining distance every hour ? then apply cross-muiltiplication.
$endgroup$
– zwim
Dec 2 '18 at 11:47
$begingroup$
Divide the total distance traveled by the total distance the two cars travel in one hour.
$endgroup$
– N. F. Taussig
Dec 2 '18 at 11:55
$begingroup$
Depending upon what direction they are traveling they may not meet but in other direction they will meet. At that time what will the impact be?
$endgroup$
– William Elliot
Dec 3 '18 at 0:30