Carlos is on his trail bike, approaching a creel bed that is 7m wide. A ramp with an incline of 10° has been built for daring people who try to jump the creek. Carlos is traveling at his bike’s maximum speed, 40km/h. a) Should Carlos attempt the jump or emphatically hit the brakes? b) What is the minimum speed a bike must have to make the jump? Assume equal elevations on either side of the creek.(adsbygoogle = window.adsbygoogle || []).push({});

Velocity = 40km/h = 11.11m/s

vx = v cos theta = 11.11m/s cos 10° =10.94m/s

vy = v sin theta = 11.11 m/s sin 10° = 1.93m/s

(Not sure for example here if I should consider significant figure on each calculation or only in the final answer)

a)

To obtain time Carlos spends in the air:

Y(t) = y0 + vy(t) -1/2(g)(t^2)

(Is it only when Y is positive upward and objects are falling downwards g should be negative?)

0m = 0m + 1.93t – (4.905m/s^2)

t(1,93m/s – (4,905m/s^2)t )

t = 0s and t =(1.93m/s)/(4.905m/s^2) = 0.40s (Again not sure if I should round or not)

To obtain distance:

X(t) = x0 + v0x(t)

X(t) = 10.94m/s(0.40s) = 4.4m and Carlos should hit the brakes!

b)

To calculate the minimum speed:

Vertical Time = 2v0y/g = 2v0 sin theta / g

And then I put time in the horizontal equation thus:

X(t) = x0 + v0x(t)

7m = 0m + v0x (2v0 sin theta / g)

7m = (v0 cos theta) (2v0 sin theta / g)

7m = (v0 cos 10) ((2v0 sin 10)/9.81)

7m = 2(v^2)(0,1666)/9,81

v^2 = 68.67/0.332 = 206.8(m/s)^2

v = 14.4m/s

Is my reasoning right? I’m really in need of your comments, and thank you in advance.

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# Homework Help: Motion in 2 and 3 dimentions

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