- #1
oooo
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This is a problem that appeared quite simple but is for some reason incorrect. I am not quite sure where I made my mistake so I spelled out my thought process below:
Question:
"Suppose that a cart was released from rest at a position of 3.5 cm down the ramp and continued to travel down the ramp to a position of 6.7 cm.The time the cart took to pass through these points is 0.2545 seconds.
What is the acceleration between these two points on the ramp?"
What I did:
I used the kinematic equation of: displacement = initial velocity* time passed + 1/2 * acceleration * (time passed )^2
I knew the initial velocity = 0 because it was released from rest and the time passed was given. I had to solve for the acceleration.
So, I converted my centimeters to meters and plugged in my values into the equation to get:
0.32 = 0*0.2545 + 1/2* a*(0.2545)^2
and then 0.32= 1/2 *a* 0.06477025
and solved for a. I got the answer of 9.88 m/s^2 approx. This, however was marked incorrect.
The question then led me to solve for the velocity at given periods of time. For example, "what is the velocity at t=0.509 seconds?" If I had a correct acceleration, wouldn't I just use the kinematic equation of acceleration= change in velocity/change in time?
Question:
"Suppose that a cart was released from rest at a position of 3.5 cm down the ramp and continued to travel down the ramp to a position of 6.7 cm.The time the cart took to pass through these points is 0.2545 seconds.
What is the acceleration between these two points on the ramp?"
What I did:
I used the kinematic equation of: displacement = initial velocity* time passed + 1/2 * acceleration * (time passed )^2
I knew the initial velocity = 0 because it was released from rest and the time passed was given. I had to solve for the acceleration.
So, I converted my centimeters to meters and plugged in my values into the equation to get:
0.32 = 0*0.2545 + 1/2* a*(0.2545)^2
and then 0.32= 1/2 *a* 0.06477025
and solved for a. I got the answer of 9.88 m/s^2 approx. This, however was marked incorrect.
The question then led me to solve for the velocity at given periods of time. For example, "what is the velocity at t=0.509 seconds?" If I had a correct acceleration, wouldn't I just use the kinematic equation of acceleration= change in velocity/change in time?