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Can anyone help me with this mechanics 1 question?
"a downhill skier travels down a slope in a straight line with constant acceleration. she passes close to four trees, T1, T2, T3, T4. she passes T2 10 secs after she passes T1 and the distance between these trees is 90m; she passes T3 15 secs after passing T2 and the distance between these trees is 172.5m. If the distance between T3 and T4 is 140m, find the time that elapses from when she passes T1 to when she passes t4 and her speed as she passes T4."
2 Answers
- Jeff AaronLv 71 decade agoFavorite Answer
Suppose the skier's speed at T1 is x, so from T1 to T2 we have:
d = ut + (1/2)at^2
90 = x(10) + (1/2)a(10)^2
90 = 10x + 50a
50a = 90 - 10x
a = (90 - 10x) / 5
a = 18 - 2x
v^2 - u^2 = 2ad
v^2 - x^2 = 2(18 - 2x)(90)
v^2 - x^2 = 3240 - 360x
v^2 = x^2 - 360x + 3240
v = sqrt(x^2 - 360x + 3240)
From T2 to T3 we have:
d = ut + (1/2)at^2
The u (initial velocity) in this section is the same as the v (final velocity) from the section above, i.e. the velocity at T2. Also, the acceleration from this section is the same as from the section above (because it never changes), i.e. it's 18 - 2x. So we have:
172.5 = sqrt(x^2 - 360x + 3240)*15 + (1/2)(18 - 2x)*15^2
172.5 = 15*sqrt(x^2 - 360x + 3240) + 2025 - 225x
172.5 - 2025 + 225x = 15*sqrt(x^2 - 360x + 3240)
225x - 1852.5 = 15*sqrt(x^2 - 360x + 3240)
Square both sides:
50625x^2 - 833625x + 3431756.25 = 225x^2 - 81000x + 729000
50625x^2 - 833625x + 3431756.25 - 225x^2 + 81000x - 729000 = 0
50400x^2 - 752625x + 2702756.25 = 0
x = (-(-752625) +/- sqrt((-752625)^2 - 4(50400)(2702756.25))) / (2*50400)
x = (752625 +/- sqrt(566444390625 - 544875660000)) / 100800
x = (752625 +/- sqrt(21568730625)) / 100800
x =~ 8.9 or 6
The latter value doesn't satisfy the original equation, so therefore:
x = (752625 + sqrt(21568730625)) / 100800 =~ 8.9
The acceleration is:
a =~ 18 - 2*8.9 =~ 0.153
The skier's velocity at T2 is:
v = sqrt(x^2 - 360x + 3240)
v =~ 56.4
The skier's velocity at T3 is:
v = u + at
v =~ 56.4 + 0.153*15.5
v =~ 58.73
Between T3 and T4 we have:
v^2 - u^2 = 2ad
v^2 = 2ad + u^2
v = sqrt(2ad + u^2)
v =~ sqrt(2*0.153*140 + 58.73^2)
v =~ 59.09 m/s
v = u + at
v - u = at
t = (v - u) / a
t =~ (59.09 - 58.73) / 0.153
t =~ 2.376
Total time elapsed from T1 to T4 is approximately 10 + 15 + 2.376 or 27.376 seconds
- 1 decade ago
My guess:
Time taken from t1 to t4 = 34s
Speed = 15.5 m/s
I used v = u +at
s = ut + .5at^2
I hope you know the above.
You can start by drawing the trees and entering the value in it.
Or i may be wrong.
Source(s): past knowledge
