Question

Jason takes off across level water on his jet-powered skis. The combined mass of Jason and his skis is 75kg (the mass of the fuel is negligible). The skis produce a forward thrust of 200N and have a coefficient of kinetic friction with water of 0.10. Unfortunately, the skis run out of fuel after only 41s. How far from his starting point has Jason traveled when he finally coasts to a stop?

Answer 1

you got any ideas yourself?

Answer 2

@IrishBoy123 Yeah my only idea so far has been 200 - (0.10*75*9.8) = 126.5 and then 126.5 = 75a and a = 1.69m/s^2. Then use the equation x = Vi(t) + 1/2at^2. Vi = 0m/s so x = 1/2(1.69m/s^2)(41^2). But that didn't work.

Answer 3

assuming you did your sums right, that's how far he travels until the engine stops.

you then need to add on the time it takes for him to stop. basically all the same calculations but without the 200N thrust.

Answer 4

Not sure how to do that.

Answer 5

do it all in reverse

this is all calculator work and i don't have a good one to hand or i'd crunch the numbers too

if i can copy and paste your work slightly, it should become apparent

for the decceleration, as friction now slows it all down:

- (0.10*75*9.8) = 75a , so a = ??.


use the equation \(v = u + at \) to find the time [v = 0] ........ or \(v^2 = u^2 + 2 a x \) to cut straight to the distance [v = 0, you know u and a]

Answer 6

@Irishboy123 What's the u equal to?

Answer 7

initial velocity

like \(v_i\) for the second part of the journey where friction is slowing it all down.

Answer 8

Yes I know that but how do I know the initial velocity?

Answer 9

first of all, the actual numbers in this don't make any sense and don't allow for reality checking. 75kg of man plus jet ski?!?!

STAGE 1 - the acceleration

your work looks good. you should get x = 1417.6m for distance travelled. using v = u + at, ypu get v = 69.15m/s for velocity at t= 41, ie when the jet ski stops working. {that's 249 kph btw!!}

STAGE 2 - the decceleration

u = 69.15 m/s {ie the previous v}, v = 0

for acceleration you now have -0.1(75)(9.8) = 75 a, a = -0.98 m/s/s

use \(v^2 = u^2 + 2 ax \implies x = \dfrac{v^2 - u^2}{2a} = \dfrac{0 - (69.1)^2}{2(-0.98)} = 2436.1\)

add that to the 1417.6 to get total distance travelled