What is the total distance travelled for an accelerating automobile?
Stephanie C asked:
An automobile accelerates from rest at 2.6m/s^2 for 18 sec. The speed is then held constant for 15 sec, after which there is an acceleration of -2.9m/s^2 until the automobile stops.
What is the total distance travelled?
This entry was posted
on Wednesday, January 6th, 2010 at 10:19 pm and is filed under Physics.
You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.
An automobile accelerates from rest at 2.6m/s^2 for 18 sec. The speed is then held constant for 15 sec, after which there is an acceleration of -2.9m/s^2 until the automobile stops.
What is the total distance travelled?
January 8th, 2010 at 5:12 pm
I dunno!
just as long as richie hammond is ok! freckles
January 10th, 2010 at 3:17 pm
R X T = D
rate times time equals distance.
you can do it! Johnny
January 10th, 2010 at 10:49 pm
x (distance)
T = Time
A = Acceleration
u = uniform speed
x = ut+1/2ATx2
That should be all you need without me giving it to you :_) Xopher
January 12th, 2010 at 1:37 am
The first distance
distance = average velocity *time
Average velocity= (initial velocity+Final velocity)/2
distance = ((0m/s + 2.6m/s)/2)*18s= 23.4m
2nd distance = average velocity * time
2nd distance = 2.6m/s * 15s = 39m
3rd distance
Initial velocity=2.6m/s
Final velocity=0m/s
acceleration= -2.9m/s^2
3rd distance = average velocity * time
average velocity = ((initial velocity + final velocity)/2)
time=(final velocity - initial velocity)/acceleration
3rd distance = ((2.6m/s+0m/s)/2) * (0m/s - 2.6m/s)/-2.9m/s^2
3rd distance = (1.3m/s)*(0.897s) =1.17m
add the three distance behroz_ahmedali
January 14th, 2010 at 12:32 am
Ok, iam assuming you know the equations of motion:
v= u + at
s=ut + 0.5at^2
v^2=u^2 + 2as
s=((u+v)/2)t
We need to break this motion up into seperate sections.
Note however that all the units in this question are in SI units, so they are consistent with each other. You don’t have kilometres which you would need to convert into metres for example, before using these equations. In that case I won’t put in the units by the numbers in my calculations everytime, just to make it look clearer. But they are there and they are consistent.
First section is from rest (u=0) accelerating at 2.6m/s^2 for 18s. We will find the displacement travelled here.
We know u(initial velocity)=0, we know a(acceleration)= 2.6 and t=18 and we want s(displacement)
the equation connecting these four is s=ut + 0.5at^2
(we don’t need v and this equation does not have v)
so s = (0 x 18) + (0.5 x 2.6 x 18^2)
= 421.2m Inviz
January 15th, 2010 at 12:57 am
Part I : The vehicle starts at zero velocity with acceleration 2.6m/s^2
so, s1=ut+1/2at^2 where u=0, s= 1/2*2.6*18*18=421.2m
Part II : At the end of part I, the vehicles velocity is v=u+at (u=0)
v=at=2.6*18=46.8m/s, now the vehicle is moving at a constant speed for 15sec, so distance(s2)=speed*time=2.6*18*15=702m.
Part III : Now the vehicle decelerates at -2.9m/s^2, and stops so final velocity is zero and initial velocity is 46.8m/s as found in part II.
v^2=u^2+2as, where v=0 and a will be negative for deceleration.
s3=46.8^2/2*2.9m
add the three distances, u get the total distance traveled by the vehicle. ANSHUL
January 17th, 2010 at 10:21 am
First part : The vehicle will have a final speed of 2.6 x 18 m/s = 46.8m/s. The distance travelled will be the average speed x time
= 46.8/2 x 18 = 421.2 metres. (the average speed will be how fast the vehicle is going at the half way point hence the division by 2)
2nd Part: The distance will be the speed x time = 46.8 x 15 = 702 metres.
3rd Part the time taken to stop will be the speed divided by the negative acceleration = 46.8/2.9 = 16.148 seconds. The distance travelled will again be the average speed x time = 46.8/2 x 16.148 = 377.86 metres.
Total distance travelled is the three distances added together. Examiner