Grade 11 acceleration problem?

  • #1

Homework Statement


Jack went up a hill to fetch water. Jack fell down and rolled down the hill accelerating at 0.65 m/s^2 down. If he rolled 11m down the hill in 3.7s, with what velocity was he moving when he fell?


Homework Equations


a = v2 - v1/t
v2 = at + v1


The Attempt at a Solution


(0.65m/s^2)(3.7s) + 11
= 13.5 m/s.

I solved for v2, the velocity when he reached the bottom of the hill.
However the answer says 1.8m/s.
What am I doing wrong? Am I supposed to be looking for something else, not v2?
Even if I solved for v1, I would get a negative answer.

As usual, I am so confused. :yuck:
 

Answers and Replies

  • #2
393
0
try this
[tex]R(t) = \frac{1}{2}at^2+v_it+R_i[/tex]
[tex]v_i = \frac{R(t) - \frac{1}{2}at^2-R_i}{t}[/tex]
 
  • #3
That confuses me even more. What does R stand for?
 
  • #4
PhanthomJay
Science Advisor
Homework Helper
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Homework Statement


Jack went up a hill to fetch water. Jack fell down and rolled down the hill accelerating at 0.65 m/s^2 down. If he rolled 11m down the hill in 3.7s, with what velocity was he moving when he fell?


Homework Equations


a = v2 - v1/t
v2 = at + v1


The Attempt at a Solution


(0.65m/s^2)(3.7s) + 11
what's that 11, that's a length unit, not a speed unit.
I solved for v2, the velocity when he reached the bottom of the hill.
However the answer says 1.8m/s.
What am I doing wrong? Am I supposed to be looking for something else, not v2?
you are supposed to be looking for the initial velocity when he fell.
Even if I solved for v1, I would get a negative answer.
and what would the negative sign imply?
 

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