Engineering Science

[david7]

If anyone can help with my last assignment question please.

A car accelerates along a straight road with a uniform acceleration when it passes A it has a speed of 12m/s and when it passes B it has a speed of 32m/s. points A and B are 1100m apart. Determine the time taken to move from A to B.

Thanks in advance

David

[vascop]

Sorry that should be "The rectangle represents the distance that would be travelled if the velocity remained at 12 m/s, and the TRIangle the distance travelled due to the increase in velocity over the time interval. "

[factor-fiction]

Given that the acceleration is constant think the easiest way to solve this in one step is to use s= (u+v)t /2.
Rearranging this gives t= 2s/(u+v)= 2x1100/(12+32)= 2200/44=50

[Old_Geezer]

Wow, that stretched some minds ;-)

[vascop]

I agree with you entirely factor. I was just trying to point out to Ignoblius where the flaw in his argument lies.
It's easy to see that he is wrong because his formula for the time difference does not depend on the initial velocity, but only on the difference between the initial and final velocities. So if the initial velocity were 1,000,000 m/s and the final velocity 1,000,020 he would get the same answer, which is obviously wrong!!

[factor-fiction]

Hi vascop- my post was in response to Ignoblius too

[roger]

Isaac must be spinning in his grave !

[Ignoblius]

vascop-You're right, I see my mistake, the area representing distance traveled up to reaching A would precede TA on the graph. The integral is the thing and that covers the rectangle and triangle areas both. Damn, I've gotten rusty at this stuff!