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Engineering Science
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
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
Answers
using the equations
v^ 2 = u^2 + 2as and v = u + at, where u=initial speed, v=final speed, a=accelerati on, s = distance travelled and t=time and rearranging,
you get
t = 2s / (v+u)
=2200/44
= 50 m
you get
t = 2s / (v+u)
=2200/44
= 50 m
23:17 Sat 23rd Jun 2012
Hi Factor. No, I didn’t mean to do that, but I did make a balls up (too much electric soup!).
Although I get the same answer as you I like to calculate it by suggesting that the average speed must be the mid point between the starting and ending rate. I would calculate that by halving the difference between the two and adding it to the lower. (32-12 =20; 20/2=10; 12+10=22). So the time needed to cover the 1100m must be 1100/22=50 seconds.
Thanks for shaking me out of my stupour. Think I’ll have another large one!
Although I get the same answer as you I like to calculate it by suggesting that the average speed must be the mid point between the starting and ending rate. I would calculate that by halving the difference between the two and adding it to the lower. (32-12 =20; 20/2=10; 12+10=22). So the time needed to cover the 1100m must be 1100/22=50 seconds.
Thanks for shaking me out of my stupour. Think I’ll have another large one!
OK,
I admit I had had a fair bit of wine when I worked out my answer last night (hence my putting 50m instead of 50s),
but the answer is fairly obviously found by dividing the distance by the average speed, as wildwood and factor30 rightly pointed out. I arrived at the same answer by a rather more circuitous route...
I admit I had had a fair bit of wine when I worked out my answer last night (hence my putting 50m instead of 50s),
but the answer is fairly obviously found by dividing the distance by the average speed, as wildwood and factor30 rightly pointed out. I arrived at the same answer by a rather more circuitous route...
My “electric soup” is not the Scottish version of milk and gas. I use the term to describe any strong-ish drinks (above 20% or thereabouts). So, not wine or beer, probably not sherry or port. To give a bit of detail which led to my brain being fuddled last night, I had drunk a fair noggin of seven star “Metaxa” which I had brought back from a recent visit to Greece. It’s a lovely drink and very moreish. So I had some more, and was incabale of dealing with the simplest of problems!
110 seconds.
Plot speed V as a function of time T. You get an up-sloping line (slope=acceleration) from (TA,12m/s) to (TB,32m/s).
The area beneath this line consists of a right triangle atop a rectangle. The rectangular area represents the distance covered up to point A, which is of no concern. The triangular area represents the distance from A to B, which is 1100m.
Area of triangle=1100m=(TB-TA)(VB-VA)/2
VB-VA=32m/s-12m/s=20m/s
DeltaT=TB-TA=(2*1100m)/(20m/s)=110s
Plot speed V as a function of time T. You get an up-sloping line (slope=acceleration) from (TA,12m/s) to (TB,32m/s).
The area beneath this line consists of a right triangle atop a rectangle. The rectangular area represents the distance covered up to point A, which is of no concern. The triangular area represents the distance from A to B, which is 1100m.
Area of triangle=1100m=(TB-TA)(VB-VA)/2
VB-VA=32m/s-12m/s=20m/s
DeltaT=TB-TA=(2*1100m)/(20m/s)=110s
-- answer removed --
Ignoblius you are wrong when you say: "The rectangular area represents the distance covered up to point A, which is of no concern.".
The rectangle represents the distance that would be travelled if the velocity remained at 12 m/s, and the rectangle the distance travelled due to the increase in velocity over the time interval.
so Area of triangle+area of rectangle=1100=(TB-TA)(VB_VA)/2+(TB-TA)VA
VB-VA=20/m/s
So DeltaT=TB-TA=1100/(20/2+12)=1100/22=50
The rectangle represents the distance that would be travelled if the velocity remained at 12 m/s, and the rectangle the distance travelled due to the increase in velocity over the time interval.
so Area of triangle+area of rectangle=1100=(TB-TA)(VB_VA)/2+(TB-TA)VA
VB-VA=20/m/s
So DeltaT=TB-TA=1100/(20/2+12)=1100/22=50