Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
primary 6 | Maths
| Speed
One Answer Below
This question need another answer.
help pls , urgent!
Date Posted: 4 years ago
Views: 803
Shorter explanation :
① In 30s, let distance moved by dog + train = 5 units (to pass each other)
Since we know they moved 1 train length in total when passing each other,
In 25s, Distance moved by John + train also also = 5 units (to pass each other)
So in 30s, this would be = 30/25 x 5 units
= 6 units
6 units - 5 units = 1 unit
So John moves more than the dog by 1 unit in every 30 seconds.
② Since train only met John 15s after passing beyond the dog ,
Distance between John and train front (when the dog is still beside the train back) = combined distance travelled in that 15s
= 15/30 x 6 units or ½ of 6 units
= 3 units
③ John's distance from the dog (after the train had passed beyond the dog)
= Distance between John and train front + distance between train front and dog(at the train back)
= 3 units + train length
= 3 units + 5 units
= 8 units
④
From ①,we know John moves more than the dog by 1 unit in every 30 seconds.
He's getting closer to the dog by 1unit every 30 seconds.
To catch the dog, John needs to catch up by 8 units of distance.
8 x 30 seconds = 240 seconds
240 seconds are needed from that point.
⑤
15 seconds after that : the train met John.
25 seconds after train met John :
the train passed beyond John.
Remaining time needed = 240 seconds - 15 seconds - 25 seconds
= 200 seconds
① In 30s, let distance moved by dog + train = 5 units (to pass each other)
Since we know they moved 1 train length in total when passing each other,
In 25s, Distance moved by John + train also also = 5 units (to pass each other)
So in 30s, this would be = 30/25 x 5 units
= 6 units
6 units - 5 units = 1 unit
So John moves more than the dog by 1 unit in every 30 seconds.
② Since train only met John 15s after passing beyond the dog ,
Distance between John and train front (when the dog is still beside the train back) = combined distance travelled in that 15s
= 15/30 x 6 units or ½ of 6 units
= 3 units
③ John's distance from the dog (after the train had passed beyond the dog)
= Distance between John and train front + distance between train front and dog(at the train back)
= 3 units + train length
= 3 units + 5 units
= 8 units
④
From ①,we know John moves more than the dog by 1 unit in every 30 seconds.
He's getting closer to the dog by 1unit every 30 seconds.
To catch the dog, John needs to catch up by 8 units of distance.
8 x 30 seconds = 240 seconds
240 seconds are needed from that point.
⑤
15 seconds after that : the train met John.
25 seconds after train met John :
the train passed beyond John.
Remaining time needed = 240 seconds - 15 seconds - 25 seconds
= 200 seconds
See 1 Answer
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
This is a challenging question for a Primary 6 student. Hope you find my solution useful!
Date Posted:
4 years ago
Click on the photo to see the answer.
abit too complicated
Guess why you didn't manage to solve the question? Because it is complicated to solve.
Applepie :
①Immediately after the train passes the dog, both are still moving.
When train has travelled past the dog, the dog is line with the end of the train.
Both the train and dog have moved a distance, and their combined distance would be equal to the length of the train.
(Eg. Let's say the train is 10m long, and 30 seconds after coming across the dog, it moves 9m.
Its front is now 9m away from the original position. If the dog didn't move, the dog will be 9m behind the train's front, and 1m away from the train's end.
But we know that the dog is now in line with the train's end now. So in this 30 seconds, the dog must have moved that 1m to reach the train's end.
Total distance the dog and train moved = 9m + 1m = 10m , which equals the train length
①Immediately after the train passes the dog, both are still moving.
When train has travelled past the dog, the dog is line with the end of the train.
Both the train and dog have moved a distance, and their combined distance would be equal to the length of the train.
(Eg. Let's say the train is 10m long, and 30 seconds after coming across the dog, it moves 9m.
Its front is now 9m away from the original position. If the dog didn't move, the dog will be 9m behind the train's front, and 1m away from the train's end.
But we know that the dog is now in line with the train's end now. So in this 30 seconds, the dog must have moved that 1m to reach the train's end.
Total distance the dog and train moved = 9m + 1m = 10m , which equals the train length
So now we know the train and dog can travel 1 train length combined, in 30 seconds.
Now the dog is beside the end of the train
In the next 15 seconds, they would travel a combined distance equal to ½ a train length. (15 ÷ 30 = ½)
Since they're going in opposite directions, the distance between them after the next 15 seconds is ½ a train length. This is where the train meets John.
The train takes 25 seconds to travel past John. In this 25 seconds, the train and dog have travelled another 5/6 of a train length combined (25 ÷ 30 = 5/6)
So now, ½ + 5/6
= 3/6 + 5/6
= 8/6
= 4/3
= 1⅓
So the end of the train and dog are 1⅓ train lengths away from each other now.
Since John is also beside the end of the train (having travelled past it in 25 seconds), John is also 1⅓ train lengths behind/away from the dog.
Now the dog is beside the end of the train
In the next 15 seconds, they would travel a combined distance equal to ½ a train length. (15 ÷ 30 = ½)
Since they're going in opposite directions, the distance between them after the next 15 seconds is ½ a train length. This is where the train meets John.
The train takes 25 seconds to travel past John. In this 25 seconds, the train and dog have travelled another 5/6 of a train length combined (25 ÷ 30 = 5/6)
So now, ½ + 5/6
= 3/6 + 5/6
= 8/6
= 4/3
= 1⅓
So the end of the train and dog are 1⅓ train lengths away from each other now.
Since John is also beside the end of the train (having travelled past it in 25 seconds), John is also 1⅓ train lengths behind/away from the dog.
Using the same logic, we can say that in 25 seconds, John and the train moved a distance of 1 train length combined.
So in 30 seconds, they will move 6/5 or 1 1/5 train lengths combined (30/25 x 1)
Compare this with our earlier info.
In 30 seconds :
distance moved by train + distance moved by John = 1 1/5 train lengths
distance moved by train + distance moved by dog = 1 train length
1 1/5 - 1 = 1/5
Since the distance moved by train in 30 seconds is the same for both cases, (constant speed)
Every 30 seconds, John moves 1/5 of a train length faster than the dog.
So he will be getting closer to the dog by 1/5 of a train length every 30 seconds.
Since John and his dog are now 1⅓ train lengths away from each other now,
1⅓ ÷ 1/5
= 4/3 x 5
= 20/3
20/3 x 30 seconds = 200 seconds
So in 30 seconds, they will move 6/5 or 1 1/5 train lengths combined (30/25 x 1)
Compare this with our earlier info.
In 30 seconds :
distance moved by train + distance moved by John = 1 1/5 train lengths
distance moved by train + distance moved by dog = 1 train length
1 1/5 - 1 = 1/5
Since the distance moved by train in 30 seconds is the same for both cases, (constant speed)
Every 30 seconds, John moves 1/5 of a train length faster than the dog.
So he will be getting closer to the dog by 1/5 of a train length every 30 seconds.
Since John and his dog are now 1⅓ train lengths away from each other now,
1⅓ ÷ 1/5
= 4/3 x 5
= 20/3
20/3 x 30 seconds = 200 seconds
@Jiachen : D isn't covered in 15 seconds by both the train and John.
15 seconds only covers the distance between John and the train's front (after travelling beyond the dog, where the dog is now beside the train's back)
But D consists of that distance plus the distance between the train front and the dog.
i.e from the train's front to the train's back, where the dog was beside. This is just the train's length.
The third equation should be
15(vj + vt) = D - L
or
15(vj + vt) + L = D
Which means D = 8/5 L and time = 240s
240s - 15s - 25s = 200s
15 seconds only covers the distance between John and the train's front (after travelling beyond the dog, where the dog is now beside the train's back)
But D consists of that distance plus the distance between the train front and the dog.
i.e from the train's front to the train's back, where the dog was beside. This is just the train's length.
The third equation should be
15(vj + vt) = D - L
or
15(vj + vt) + L = D
Which means D = 8/5 L and time = 240s
240s - 15s - 25s = 200s
@J Your are right. I was careless.
Still, great effort and kudos to you for providing a logical algebraic method that is well presented. (albeit the small error).
On a side note, I guess we have to treat the dog and John's 'length' as negligible for the question. Not sure how it would turn out if those were considered.
On a side note, I guess we have to treat the dog and John's 'length' as negligible for the question. Not sure how it would turn out if those were considered.
360 Tutoring Program