Asked by GRISSELL
Please help and let me know if I did something wrong.
If a cart is released from rest at a 10 meter incline and takes 28 seconds to travel down this incline.
1. Calculate the average speed of the cart? I wrote s=ad/at=10m/28 sec = 357 m/s
2. calculate the instantaneous velocity of the cart at the end of the 10 meters(this value should be twice the value in question 1 . I wrote the answer .714
3. calculate the acceleration of the cart. I wrote a=delta v/delta t=.357/28 seconds answer is a=.01275 m/s^2
4. how long does it take the cart to travel the first 5 meters down the track? (the answer is not 14 seconds)
5=0+0.1275 (t)
5-.01275
answer is t=4.98725 m
If a cart is released from rest at a 10 meter incline and takes 28 seconds to travel down this incline.
1. Calculate the average speed of the cart? I wrote s=ad/at=10m/28 sec = 357 m/s
2. calculate the instantaneous velocity of the cart at the end of the 10 meters(this value should be twice the value in question 1 . I wrote the answer .714
3. calculate the acceleration of the cart. I wrote a=delta v/delta t=.357/28 seconds answer is a=.01275 m/s^2
4. how long does it take the cart to travel the first 5 meters down the track? (the answer is not 14 seconds)
5=0+0.1275 (t)
5-.01275
answer is t=4.98725 m
Answers
Answered by
drwls
1. 10/28 is not 357. You left out a decimal point.
2. Correct
3. Correct
4. It takes 10/sqrt2 seconds to go the first half of the distance, because distance travelled is proportional to t^2.
I do not know what your question#5 is. It looks like you are subtracting an acceleration from a distance, which makes no sense
2. Correct
3. Correct
4. It takes 10/sqrt2 seconds to go the first half of the distance, because distance travelled is proportional to t^2.
I do not know what your question#5 is. It looks like you are subtracting an acceleration from a distance, which makes no sense
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