M1 Maths HELP

A particle accelerates from rest with an acceleration of 3ms^-2 to speed V. It continues at this speed for time T then decelerates to rest at 1.5ms ^-2.
The total time for this motion is 60 seconds and the total distance travelled is 1000 m, find the value of V.

I've tried so many method and they're not working :frown:

Reply 1

Jozanic

OP

13

Original post by jordanshelley97

You can use SUVAT.

S = 1000
U = 0
V = ?
A = 3
T = ?

We can use v^2 = u^2 + 2as
v^2 = 6000
v = square root of 6000

Isn't that incorrect because it doesn't have an acceleration of 3ms^-2 throughout the whole motion

Original post by Jozanic

Isn't that incorrect because it doesn't have an acceleration of 3ms^-2 throughout the whole motion

Draw a speed-time graph and it can help alongside SUVAT. The graphs has 3 parts where you can apply SUVAT - firstly when it's accelerating at 3m/s^2, then when it's not accelerating, and finally when it is decelerating.

The area under this graph should be 1000m.

Original post by jordanshelley97

You can use SUVAT.

S = 1000
U = 0
V = ?
A = 3
T = ?

We can use v^2 = u^2 + 2as
v^2 = 6000
v = square root of 6000

Acceleration isn't constant throughout the entire 1000m therefore it doesn't work the way you did it.

Reply 4

Jozanic

OP

13

Original post by RDKGames

Draw a speed-time graph and it can help alongside SUVAT. The graphs has 3 parts where you can apply SUVAT - firstly when it's accelerating at 3m/s^2, then when it's not accelerating, and finally when it is decelerating.

The area under this graph should be 1000m.

I did exactly that but I'm still getting weird answers.

Original post by Jozanic

I did exactly that but I'm still getting weird answers.

Show your working.

Reply 6

ManLike007

19

You could draw a speed time graph. The area under the graph is the total distance travelled.

Reply 7

Jozanic

OP

13

Original post by RDKGames

Show your working.

AB (Accelerating)
s =
u = 0
v = V
a = 3
t =

BC (Constant)
s =
u = V
v = V
a = 0
t = T

CD (Decelerating)
s =
u = V
v = 0
a = -1.5
t =

I made S the subject using v^2 = u^2+2as and tried to equate the accelerating and decelerating part.

(edited 7 years ago)

Original post by Jozanic

AB (Accelerating)
s =
u = 0
v = V
a = 3
t =

BC (Constant)
s =
u = V
v = V
a = 0
t = T

CD (Decelerating)
s =
u = V
v = 0
a = -1.5
t =

I made S the subject using v^2 = u^2+2as and tried to equate the accelerating and decelerating part.

Try the following:

1000=\frac{1}{2}(Vt_1)+V(t_2-t_1)+\frac{1}{2}V(60-t_2)

and

\displaystyle \frac{V}{t_1}=3

and

\displaystyle \frac{-V}{60-t_2}=-1.5

You can easily get this from drawing the graph and labeling the end of first interval as

t_1

, end of second as

t_2

and end of third as 60.

(edited 7 years ago)

Reply 9

Jozanic

OP

13

Original post by RDKGames

Try the following:

1000=\frac{1}{2}(Vt_1)+V(t_2-t_1)+\frac{1}{2}V(60-t_2)

and

\displaystyle \frac{V}{t_1}=3

and

\displaystyle \frac{-V}{60-t_2}=-1.5

You can easily get this from drawing the graph and labeling the end of first interval as

t_1

, end of second as

t_2

and end of third as 60.

Ahhhh thanks. I sort of get the idea now. You're adding the separate parts of the graph. But what is the second section with V/t1 = 3 and -V/(60/t2) =-1.5

Original post by Jozanic

Ahhhh thanks. I sort of get the idea now. You're adding the separate parts of the graph. But what is the second section with V/t1 = 3 and -V/(60/t2) =-1.5