How Far Will the Cart Travel at Increased Speeds?

[Ahmad Hossain]

First read the question....I am mechanical engineer and I am working with motion...so he asked us that A cart has initial velocity is 0.5m/s and displacement 1m and the same track, the experiment do again and the cart velocity 1m/s and how far it moves? he mentions same track and the experiment do again...that is the mean key to understand the problem...in this question he only wants to know the displacement...first find out its time when it moves 1m...So v=X/t...0.5=1/t and t=2s... Now its initial velocity 1m/s and inital velocity can differ and it depends on your applied force...so you can travel same distance in a same time and initial velocity can differ. final answer x=vt...x=1*2 =2m

 

[Doc Al]

first find out its time when it moves 1m...So v=X/t...0.5=1/t and t=2s...

This calculation is incorrect. The speed is not constant. (Hint: Find the average speed.)

Now its initial velocity 1m/s and inital velocity can differ and it depends on your applied force...so you can travel same distance in a same time and initial velocity can differ. final answer x=vt...x=1*2 =2m

This is wrong. (For multiple reasons, including the same error made above.) You cannot just assume the the same time. Use physics!

 

[Ahmad Hossain]

I think it is best you let someone else help on this one. You are only telling me I do not understand physics instead of solving the question correctly. I already provided you the answer and then you just told me my answer again.

Not necessarily. The answer key can be wrong.

The key is: what is the same between the two trails? The assumption must be that the retarding force and thus the acceleration is the same.

Which leads to your answer of 4 m. The faster car has 4 times the energy and thus 4 times the work must be done to stop it and thus 4 times the distance.

EDIT: Sorry, posted my mistake. Can't figure out how to delete it!

what you do not understand.

This calculation is incorrect. The speed is not constant. (Hint: Find the average speed.)This is wrong. (For multiple reasons, including the same error made above.) You cannot just assume the the same time. Use physics!

I use same time because of his question and you can change initial velocity because you applied force is different.the box has same mass and the track is same...it is better you do the experiment. I am sure 100%...

 

[Doc Al]

I use same time because of his question and you can change initial velocity because you applied force is different.the box has same mass and the track is same...

What do you mean by "applied force"? The force that gives the car its initial speed? That is irrelevant. What matters is the force from the track, which is the same.

I am sure 100%...

But still wrong.