What is the total distance travelled for an accelerating automobile?
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?
Answers:
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 << note this displacement, we're going to use it at the end
The next section, we have the automobile continuing its motion at a constant velocity that it has gained after accelerating at 2.6m/s^2 for 18s.
Here, we need to find what this constant velocity is, which is easy.
Since the body is accelerating at 2.6 metres per second every second it is gaining a velocity of 2.6 metres per second, every second.
So the velocity it has gained after it has accelerated for 18s is just
2.6m/s^2 multiplied by 18s
= 46.8m/s^1
It's important you notice that I am using 18s because this is how much time it has accelerated for to reach that constant velocity; it stopped accelerating after 18s. 15s is the time it travels at that speed it has gained after accelerating for 18s.
Now this velocity is constant so every second it travels a displacement of 46.8m, the displacement it travels here is just 46.8m/s^1 multiplied by 15s
= 702m << note this as well.
We have one last section now. Here, it is accelerating at -2.9m/s^2 from a velocity of 46.8m/s^1 (the constant velocity we calculated remember) to a velocity of 0m/s^1 (rest). So we need one of the equations of motion here.
We know u=46.8m/s^1, a= -2.9m/s^2 and v must = 0m/s^2
we want s. No t is involved here so the equation that doesn't have t is
v^2=u^2 + 2as
This equation isn't giving us displacement in this form though. We need to rearrange to make s the subject by leaving it alone on one side of the equal sign.
First lets put u^2 to the other side of the equal sign (iam taking it away from both sides of the equal sign to do this).
That gives us v^2 - u^2=2as
Now we just need to divide both sides by 2a to leave s on its own.
That gives us
(v^2 - u^2)/2a = s
Now referring back to our values, we just subsitute them in,
which gives us
(0^2 - 46.8^2) / (2 x -2.9) = 377.6275862m << note this lastly
Remember all them displacements that we noted, now all we have to do now is add them together. To find the total displacement travelled. That is
421.2m + 702m + 377.6275862m = 1500.827586m
Which is your answer
A few notes here:
The equations we used gives us displacement only, not distance travelled. That's why I always referred to them as displacements. And same for velocity. The equations only deal with velocity, not speed. The difference is that speed doesn't care about it's direction, velocity does. However, the automobile just travels in a straight line and straight forward in this case; there isn't any case where it travels backwards or anything. It just travels in one direction, so displacement automatically gives us distance travelled anyway and velocities in this case is the same for speed.
Also I didn't round up the any numbers (especially that long 377.62. number) becuase I don't want to round numbers until I get to the final answer, becuase that will introduce inaccuracies. I can however round up now at the final answer.
I leave that bit to you :)
And also, yes I know I showed this like I would show it to a primary school kid. Not coz I think you are one, just to make it as clear as possible.
I dunno!
just as long as richie hammond is ok!
R X T = D
rate times time equals distance.
you can do it!
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 :_)
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
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.
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.
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.
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What is the total distance travelled?
Answers:
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 << note this displacement, we're going to use it at the end
The next section, we have the automobile continuing its motion at a constant velocity that it has gained after accelerating at 2.6m/s^2 for 18s.
Here, we need to find what this constant velocity is, which is easy.
Since the body is accelerating at 2.6 metres per second every second it is gaining a velocity of 2.6 metres per second, every second.
So the velocity it has gained after it has accelerated for 18s is just
2.6m/s^2 multiplied by 18s
= 46.8m/s^1
It's important you notice that I am using 18s because this is how much time it has accelerated for to reach that constant velocity; it stopped accelerating after 18s. 15s is the time it travels at that speed it has gained after accelerating for 18s.
Now this velocity is constant so every second it travels a displacement of 46.8m, the displacement it travels here is just 46.8m/s^1 multiplied by 15s
= 702m << note this as well.
We have one last section now. Here, it is accelerating at -2.9m/s^2 from a velocity of 46.8m/s^1 (the constant velocity we calculated remember) to a velocity of 0m/s^1 (rest). So we need one of the equations of motion here.
We know u=46.8m/s^1, a= -2.9m/s^2 and v must = 0m/s^2
we want s. No t is involved here so the equation that doesn't have t is
v^2=u^2 + 2as
This equation isn't giving us displacement in this form though. We need to rearrange to make s the subject by leaving it alone on one side of the equal sign.
First lets put u^2 to the other side of the equal sign (iam taking it away from both sides of the equal sign to do this).
That gives us v^2 - u^2=2as
Now we just need to divide both sides by 2a to leave s on its own.
That gives us
(v^2 - u^2)/2a = s
Now referring back to our values, we just subsitute them in,
which gives us
(0^2 - 46.8^2) / (2 x -2.9) = 377.6275862m << note this lastly
Remember all them displacements that we noted, now all we have to do now is add them together. To find the total displacement travelled. That is
421.2m + 702m + 377.6275862m = 1500.827586m
Which is your answer
A few notes here:
The equations we used gives us displacement only, not distance travelled. That's why I always referred to them as displacements. And same for velocity. The equations only deal with velocity, not speed. The difference is that speed doesn't care about it's direction, velocity does. However, the automobile just travels in a straight line and straight forward in this case; there isn't any case where it travels backwards or anything. It just travels in one direction, so displacement automatically gives us distance travelled anyway and velocities in this case is the same for speed.
Also I didn't round up the any numbers (especially that long 377.62. number) becuase I don't want to round numbers until I get to the final answer, becuase that will introduce inaccuracies. I can however round up now at the final answer.
I leave that bit to you :)
And also, yes I know I showed this like I would show it to a primary school kid. Not coz I think you are one, just to make it as clear as possible.
I dunno!
just as long as richie hammond is ok!
R X T = D
rate times time equals distance.
you can do it!
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 :_)
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
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.
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.
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.
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