So the original problem reads:
"A 450 kg mass is accelerated at 2.5 m/s² : a) what is the force causing this acceleration? b) how fast is the mass traveling at the end of 3.0 seconds? c) how much distance will have been covered after 3.5 seconds?"
A and B were pretty simple for me, I got a force of 1125 Newtons and the velocity after 3 sec would be 7.5 m/s (we have an answer key and these answers checked out).
However, I thought I had C down, simply solving d = vt (v was 7.5 and t was 3.5 sec) and I first got an answer of 26.25 meters. But then I checked the key and saw that it was wrong so I tried using the equation d = v0t + (1/2)at² and ended up with an answer of 41.56 meters, also WRONG.
The answer the key gives is 15.3 m but I couldn't figure out how to get this answer. So I searched the internet and found a step by step solution guide for the same exact problem and it uses the equation x = (1/2)at² and does indeed get the correct answer of 15.3 m.
Why does it use the equation x = (1/2)at² ? I thought that that equation was primarily used just for free falling objects, which this problem is not.
Please explain to me.
"A 450 kg mass is accelerated at 2.5 m/s² : a) what is the force causing this acceleration? b) how fast is the mass traveling at the end of 3.0 seconds? c) how much distance will have been covered after 3.5 seconds?"
A and B were pretty simple for me, I got a force of 1125 Newtons and the velocity after 3 sec would be 7.5 m/s (we have an answer key and these answers checked out).
However, I thought I had C down, simply solving d = vt (v was 7.5 and t was 3.5 sec) and I first got an answer of 26.25 meters. But then I checked the key and saw that it was wrong so I tried using the equation d = v0t + (1/2)at² and ended up with an answer of 41.56 meters, also WRONG.
The answer the key gives is 15.3 m but I couldn't figure out how to get this answer. So I searched the internet and found a step by step solution guide for the same exact problem and it uses the equation x = (1/2)at² and does indeed get the correct answer of 15.3 m.
Why does it use the equation x = (1/2)at² ? I thought that that equation was primarily used just for free falling objects, which this problem is not.
Please explain to me.
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Two things you need to realize:
The equation
(1) d = v0t + (1/2)at²
contains the term v0 * t. What did you use for v0?
v0--also referred to as u or Vi--is the initial velocity. In other words, the velocity in effect at 0 seconds. Since the problem doesn't state it, you have to assume that it's zero.
Substituting v0 = 0 in (1) gives you the equation you used to get the correct answer.
Second, the equation
(2) x = (1/2)at²
works for ANY kind of acceleration, not just gravity's, which is 9.8m/s^2 directed toward the center of the earth. So you must also use it for this problem.
Hope this clarifies those two Equation of Motion issues for you.
.
The equation
(1) d = v0t + (1/2)at²
contains the term v0 * t. What did you use for v0?
v0--also referred to as u or Vi--is the initial velocity. In other words, the velocity in effect at 0 seconds. Since the problem doesn't state it, you have to assume that it's zero.
Substituting v0 = 0 in (1) gives you the equation you used to get the correct answer.
Second, the equation
(2) x = (1/2)at²
works for ANY kind of acceleration, not just gravity's, which is 9.8m/s^2 directed toward the center of the earth. So you must also use it for this problem.
Hope this clarifies those two Equation of Motion issues for you.
.
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Thanks, Alexander; however, I see that I did miss one of your points:
" . . . I thought I had C down, simply solving d = vt (v was 7.5 and t was 3.5 sec) . . . "
Using that, you're saying that the velocity is uniform for the 3.5s. But it isn't, since it starts slower and finishes faster than 7.5.
" . . . I thought I had C down, simply solving d = vt (v was 7.5 and t was 3.5 sec) . . . "
Using that, you're saying that the velocity is uniform for the 3.5s. But it isn't, since it starts slower and finishes faster than 7.5.
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