Decceleration problem with 3 stages of motion

[oyster21]

Homework Statement

A vehicle of a mass 1000kg performs 3 stages of motion which are:

1)Accelerates uniformly from rest to a velocity of 40m/s in 5 seconds.

2)Continues at this velocity for 10 seconds

3)decelerates uniformly at a rate of 3m/s[SUP]2[/SUP]

Relevant Equations

V= U+at > a=(V-U)/t

1):
a=(V-U)/t
a=(40m/s - 0m/s) /5sec = 8m/s²

2):
U-V/t > t= V/a
t= 0m/s - 40m/s / -3m/s² = 13.3sec

3):
8m/s² = 8x8= 64m
64m x 10sec = 640m

How do I work out the total distance traveled over the 3 stages with the information if its correct for question 3?

and do I have to take into consideration the mass 1000kg?

 

[haruspex]

You have not quoted the questions posed. I assume the first is to find the acceleration in the first stage and the second to find the time to come to rest in the third stage.
What is the third question? Is it to find the total distance? Average speed?
8m/s2 = 8x8= 64m and 64m x 10sec = 640m make no sense.

 

[oyster21]

You have not quoted the questions posed. I assume the first is to find the acceleration in the first stage and the second to find the time to come to rest in the third stage.
What is the third question? Is it to find the total distance? Average speed?
8m/s2 = 8x8= 64m and 64m x 10sec = 640m make no sense.

A vehicle of a mass 1000kg performs 3 stages of motion which are:

1)Accelerates uniformly from rest to a velocity of 40m/s in 5 seconds.

2)Continues at this velocity for 10 seconds

3)decelerates uniformly at a rate of 3m/s²

Question:

1) find the acceleration in the1st stage?

2)the time taken to come to rest?

3) the overall distance travelled?I have tried to answer Q3 but only partly as I don’t know how to answer it.
I was thinking I had to to take time it took to accelerate which is 8m/s² then do I use that and say 8m/s² = 8x8 = 64m then times that by 10 seconds giving me 640m and then work out the distance it took to comt to rest?

as you can probably tell I need some help with working out the answer so any guidance or input is appreciated.

 

[haruspex]

time it took to accelerate which is 8m/s2

No, the time it took to accelerate is 5 seconds. Did you mean the time it took to accelerate to 8m/s2?

say 8m/s2 = 8x8 = 64m

As I posted, that makes no sense. How can m/s² become just m? Why are you squaring the acceleration ? What standard SUVAT equation do you think you are applying?

 

[oyster21]

Ok
My answer to the 1st question is the acceleration is 8m/s²

my answer to the 2nd question is time taken to come to rest is 13.3seconds

how would I work out the 3rd question?

also can you tell me if my answers so far are correct?

thank you for your patience though.

 

[Lnewqban]

...
as you can probably tell I need some help with working out the answer so any guidance or input is appreciated.

If you could construct a graph, many things would become clear.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/motgraph.html#c1

 

[haruspex]

how would I work out the 3rd question?

Start by explaining the thinking behind the attempt you made:

8m/s2 = 8x8= 64m
64m x 10sec = 640m

What standard SUVAT equations do you think you are applying?

 

[oyster21]

Start by explaining the thinking behind the attempt you made:

What standard SUVAT equations do you think you are applying?

u = 0
t = 28.3sec (combined time of all 3 stages)
a = 8
s = ?

S= ut+½at²

= 0 + ½ (8)(28.3)²

= 3,203.56m (total distance travelled)

is this correct?

 

[haruspex]

u = 0
t = 28.3sec (combined time of all 3 stages)
a = 8
s = ?

S= ut+½at²

= 0 + ½ (8)(28.3)²

= 3,203.56m (total distance travelled)

is this correct?

You seem to have calculated the distance it would have covered if it had maintained its original acceleration for the whole 28.3 seconds.
The SUVAT equations assume constant acceleration. You can only apply them over an interval during which the acceleration does not change. There are three intervals in this question with different accelerations, so you cannot apply one equation over more than one of the intervals.
Work out the distance covered in each interval separately.

 

[oyster21]

You seem to have calculated the distance it would have covered if it had maintained its original acceleration for the whole 28.3 seconds.
The SUVAT equations assume constant acceleration. You can only apply them over an interval during which the acceleration does not change. There are three intervals in this question with different accelerations, so you cannot apply one equation over more than one of the intervals.
Work out the distance covered in each interval separately.

would i use the same equation to work out each stage seperatley?

 

[oyster21]

You seem to have calculated the distance it would have covered if it had maintained its original acceleration for the whole 28.3 seconds.
The SUVAT equations assume constant acceleration. You can only apply them over an interval during which the acceleration does not change. There are three intervals in this question with different accelerations, so you cannot apply one equation over more than one of the intervals.
Work out the distance covered in each interval separately.

i've used the same equations for each stage. ( s=ut+½at²)
1st stage distance = 100m
2nd stage distance= 797.3m
3rd stage distance = 400m

100m+797.3m+400m = 1297.3m

?

 

[haruspex]

i've used the same equations for each stage. ( s=ut+½at²)
1st stage distance = 100m
2nd stage distance= 797.3m
3rd stage distance = 400m

100m+797.3m+400m = 1297.3m

?

The 100m is right, but the other two are wrong.
Please post your working.

 

[Lnewqban]

u = 0
t = 28.3sec (combined time of all 3 stages)
a = 8
s = ?

S= ut+½at²
...

Have you been able to understand the question to the point of being confident with your use of that equation?

Again, if you make a graph of velocity versus time, you could easily see the whole process and the details of each of the three stages.
The area between the curve and the time-axis always equals the total distance covered by the vehicle.

Stage 1:
The first term of the equation becomes zero, because the vehicle accelerated from repose.

Stage 2:
The second term of the equation becomes zero, because the vehicle had no acceleration.

Stage 3:
The first term of the equation would be the distance to be covered if sustaining the speed of 40 m/s for the duration time of this stage 3, which must be reduced by subtracting the second term from it, because the vehicle is on negative acceleration down to repose.
This stage is the reverse of stage 1, but with a smaller rate of acceleration (negative in this case).

 

[oyster21]

The 100m is right, but the other two are wrong.
Please post your working.

Stage 1:
Acceleration from rest to 40m/s.
Distance travelled
S=ut+^1/2at²
S=0x5+0.5(8)(5)²=100m

Travelled at 40m/s over 10seconds
Distance travelled
S=ut+^1/2at²
S=40x10+0.5(0)(10)²=400m

stage 3:
Deceleration from 40/ms at 3m/s²
Distance travelled
S=ut+^1/2at²
S=40x13.3+0.5x(3)(13.3)²=797.3

total distance:
100m+400m+797.3m=1297.3m

where am I going wrong?

 

[oyster21]

Have you been able to understand the question to the point of being confident with your use of that equation?

Again, if you make a graph of velocity versus time, you could easily see the whole process and the details of each of the three stages.
The area between the curve and the time-axis always equals the total distance covered by the vehicle.

Stage 1:
The first term of the equation becomes zero, because the vehicle accelerated from repose.

Stage 2:
The second term of the equation becomes zero, because the vehicle had no acceleration.

Stage 3:
The first term of the equation would be the distance to be covered if sustaining the speed of 40 m/s for the duration time of this stage 3, which must be reduced by subtracting the second term from it, because the vehicle is on negative acceleration down to repose.
This stage is the reverse of stage 1, but with a smaller rate of acceleration (negative in this case).

View attachment 270494

Is you graph correct as it stayed at 40m/s for 10seconds

 

[PeroK]

stage 3:
Deceleration from 40/ms at 3m/s²
Distance travelled
S=ut+^1/2at²
S=40x13.3+0.5x(3)(13.3)²=797.3

total distance:
100m+400m+797.3m=1297.3m

where am I going wrong?

You didn't notice that the vehicle is slowing down during phase 3.

 

[oyster21]

You didn't notice that the vehicle is slowing down during phase 3.

stage 3:
Deceleration from 40/ms at 3m/s²
Distance travelled
S=ut+1/2at²
S=40x13.3-0.5x(3)(13.3)²=265.3

is that rite for stage 3?

 

[PeroK]

stage 3:
Deceleration from 40/ms at 3m/s²
Distance travelled
S=ut+1/2at²
S=40x13.3-0.5x(3)(13.3)²=265.3

is that rite for stage 3?

Looks about right. Just under $14s$ at an average speed of $20m/s$.

 

[oyster21]

Looks about right. Just under $14s$ at an average speed of $20m/s$.

so the total distance traveled would be:

100m+400m+265.3m=765.3m

 

[PeroK]

so the total distance traveled would be:

100m+400m+265.3m=765.3m

Looks about right.

I get $766.7m$.

 

[haruspex]

where am I going wrong?

Glad to see you have it right now. You confused me in post #11 by switching your answers for stages 2 and 3.

 

[oyster21]

Glad to see you have it right now. You confused me in post #11 by switching your answers for stages 2 and 3.

i confused myself.. thanks for your earlier input though.