This preview shows page 1. Sign up to view the full content.
Solution
TAM 212
Sample Exam number 1, Chapter 1; Monday Sept 13, 2004.
1]
A particle increases its speed along its path at a uniform rate of 1m/sec
2
from its original position at rest
at point A.
Its path is straight for 8 meters and parabolic thereafter.
Find the speed at B.
Also find the magnitude of its acceleration vector 
a
 immediately
after
entering the curved section
at B.
Note that its speed is still increasing after the particle passes B.
Answer:
8 meters = (1/2) a t
2
; s=with a = 1 so t = 4 seconds and its speed then is 4 m/sec when it
gets to point B.
[ alternatively you could write the familiar(?) v = sqrt(2a d) ]
Just
after it passes B it is moving horizontally with (now we use path variables) sdot =
4m/sec, sdoubledot = 1m/sec
2
(recall it is still increasing its speed) and rho = radius of
curvature = (1+y’
2
)
3/2
/y” = 10 meters.
So
a
= 1
e
t
+ (4)
2
/10
e
n
= 1
e
t
+ 1.6
e
n
.
And 
a
 =
sqrt(1+1.6
2
) = 1.89 m/sec
2
.
2]
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '10
 richard

Click to edit the document details