# Homework Help: Motion in 2 and 3 dimentions

1. Feb 18, 2004

### Beretta

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.

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

2. Feb 18, 2004

### HallsofIvy

Looks good to me. As for your specific questions:
"(Not sure for example here if I should consider significant figure on each calculation or only in the final answer)"
As a general rule, include one more figure in your calculations than you put in your answer. In this case, you are given, at best, 2 significant figures so you need 2 in your answer.

"(Is it only when Y is positive upward and objects are falling downwards g should be negative?)"
A coordinate system is something you impose on the problem yourself. What most people do- and I recommend- is take y to be positive upward (so "falling down" is negative) and g negative.

You calculated the distance Carlos will go before "touching down" again and found that it was less than the distance across the creek.
Okay, he shouldn't try it. I see no problem with your calculation.

You finally calculated the speed necessary so that horizontal dstance is at least 7 m. Once again, it looks good to me.

Finally, you calculated the minimum speed necessary to clear the creek and it was slightly MORE than Carlo's actual speed.

3. Feb 18, 2004

### Beretta

Thank you very much, I appreciate your comment.

Last edited: Feb 18, 2004