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M1 speed time graphs

A car starts from rest and moves with constant acceleration along a straight horizontal road. The car reaches a speed of V

ms^{-1}

in 20 secs. It moves at a constant speed V

ms^{-1}

for the next 30 secs, then moves with constant deceleration 0.5

ms^{-2}

. It moves at speed 8

ms^{-1}

for the next 15 secs and then moves with constant deceleration 1/3

ms^{-2}

until lit comes to rest.

Sketch a speed time graph for this journey
Find
the value of V
total time for the journey
total distance travelled by the car

I've done the graph and found V
I got V as 14

ms^{-1}

i've split up the graph after 50s into a trapezium rectangle and triangle in that order

Edit:actually nvm it's just usiing v=u+at a bunch xD sorry just disregard this thread until i get to part D(distance bit)

(edited 7 years ago)

Reply 1

Zacken

Answering to bump off the unanswered list.

Reply 2

Steelmeat

Original post by Zacken

Answering to bump off the unanswered list.

For the total distance i got 908 Metres is that right?

Reply 3

Zacken

Original post by Steelmeat

For the total distance i got 908 Metres is that right?

I dunno, post your working.

Reply 4

Steelmeat

Original post by Zacken

I dunno, post your working.

oh boy this is gonna be long one

Reply 5

Steelmeat

Original post by Zacken

I dunno, post your working.

so we need to do all parts of the question besides sketch the graph to know the answer for the final question

For V
v=v u=0 s=140 t=20

s=\left( \dfrac{v+u}{2}\right)t

140=10v
v=14

ms^{-1}

for total time
the time for the first chunk of the journey we're told is 50secs
we calculate the time for the last 2 remaining section on the graph
u=14 v=8 a=-0.5 t=t
v=u+at
8=14-0.5t
0.5t=6
t=12

u=8 v=0 a=

-\frac{1}{3}

t=t
v=u+at
0=8-

\frac{1}{3}

t
t=24

50+12+15+24=101 secs

distance i separated my graph up into shapes
from left to right
a triangle: 0.5x20x16= 140
a rectangle: 14x30=420
a trapezium: 0.5(14+8)12=132
a rectangle: 15x8=120
a thin triangle: 0.5x8x24=96

140+420+132+120+96=908

so is it right or did i make a huge mistake somewhere?

(edited 7 years ago)

Reply 6

Steelmeat

@notnek
is it right? ^^

Reply 7

Notnek

Original post by Steelmeat

It moves at a constant speed V

ms^{-1}

for the next 30 secs, then moves with constant deceleration 0.5

ms^{-2}

. It moves at speed 8

ms^{-1}

for the next 15 secs...

This part of your question is confusingly worded. Have you posted the question as it is written?

Reply 8

Steelmeat

Original post by notnek

This part of your question is confusingly worded. Have you posted the question as it is written?

let me check that bit again ...

sorry my bad i missed a bit out
It moves at constant speed V

ms^{-1}

for the next 30 seconds, then moves with constant deceleration 0.5

ms^{-2}

until it has speed 8

ms^{-1}

Reply 9

Zacken

Original post by Steelmeat

let me check that bit again ...

sorry my bad i missed a bit out
It moves at constant speed V

ms^{-1}

for the next 30 seconds, then moves with constant deceleration 0.5

ms^{-2}

until it has speed 8

ms^{-1}

Just post the full question as a picture or link.

Reply 10

Steelmeat

Original post by Zacken

Just post the full question as a picture or link.

ok
a picture it is then

Edit: sorry about that rest your necks with this

(edited 7 years ago)

DSC_0007[1].jpg505.8KB

Reply 11

Zacken

Original post by Steelmeat

140+420+132+120+96=908

so is it right or did i make a huge mistake somewhere?

After numerous neck craning: that's right.

Reply 12

Steelmeat

Original post by Zacken

After numerous neck craning: that's right.

haha sorry about that i realised it was upside down and unfortunately you came to early and i hadn't posted the edited version yet sorry about that xD

Reply 13

Zacken

Original post by Steelmeat

haha sorry about that i realised it was upside down and unfortunately you came to early and i hadn't posted the edited version yet sorry about that xD