at time t=0 particle is projected with a velocity of 2 i m/s from point with position vector ( 10i + 90j ) m/s . the position ve
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We can set it up like this, where <em>s </em>is the speed of the canoeist:
To make a common denominator between the fractions, we can multiply the whole equation by s(s-5):
![s(s-5)[\frac{18}{s} + \frac{4}{s-5} = 3] \\ 18(s-5)+4s=3s(s-5) \\ 18s - 90+4s=3 s^{2} -15s](https://tex.z-dn.net/?f=s%28s-5%29%5B%5Cfrac%7B18%7D%7Bs%7D%20%2B%20%5Cfrac%7B4%7D%7Bs-5%7D%20%3D%203%5D%20%5C%5C%2018%28s-5%29%2B4s%3D3s%28s-5%29%20%5C%5C%2018s%20-%2090%2B4s%3D3%20s%5E%7B2%7D%20-15s)
If we rearrange this, we can turn it into a quadratic equation and factor:
Technically, either of these solutions would work when plugged into the original equation, but I would use the second solution because it's a little "neater." We have the speed for the first part of the trip (9 mph); now we just need to subtract 5mph to get the speed for the second part of the trip.
The canoeist's speed on the first part of the trip was 9mph, and their speed on the second part was 4mph.
To make a common denominator between the fractions, we can multiply the whole equation by s(s-5):
If we rearrange this, we can turn it into a quadratic equation and factor:
Technically, either of these solutions would work when plugged into the original equation, but I would use the second solution because it's a little "neater." We have the speed for the first part of the trip (9 mph); now we just need to subtract 5mph to get the speed for the second part of the trip.
The canoeist's speed on the first part of the trip was 9mph, and their speed on the second part was 4mph.