Need help determining distance from acceleration (1 Viewer)

[pi-ka-chew]

A car moves along a straight road from its front gate, where it is initially stationary. During the first 10s, it has a constant acceleration of 2m/s^2, it has zero acceleration during the next 30 seconds, and it decelerates at 1m/s^2 for the final 20 s.

how far does the car go altogether?

so i said that:
during first 10s, v = 2t
during next 30s, v = c (where c is a constant, and i'm not sure if this works, but since in the 10th second, it reaches a velocity of 20m/s, this c = 20)
during last 20s, v = -t

i guess i did this the long way, but i plugged in t = 1,2...,10 and obtained the velocities and hence distance. e.g. when t = 1, v = 1, so d =1, when t = 2, v=4 so d = 4. i'm not sure if i'm doing this correctly. the final answer i got is 110 (first 10s)+ 600(next 30) + 190(last 20) = 900

and althought the answer is 900, but based on the answers, it should be 100+600+200 = 900.

*note: physics formulae cannot be used*

thanks for your help!

 

[random-1006]

A car moves along a straight road from its front gate, where it is initially stationary. During the first 10s, it has a constant acceleration of 2m/s^2, it has zero acceleration during the next 30 seconds, and it decelerates at 1m/s^2 for the final 20 s.

how far does the car go altogether?

so i said that:
during first 10s, v = 2t
during next 30s, v = c (where c is a constant, and i'm not sure if this works, but since in the 10th second, it reaches a velocity of 20m/s, this c = 20)
during last 20s, v = -t

i guess i did this the long way, but i plugged in t = 1,2...,10 and obtained the velocities and hence distance. e.g. when t = 1, v = 1, so d =1, when t = 2, v=4 so d = 4. i'm not sure if i'm doing this correctly. the final answer i got is 110 (first 10s)+ 600(next 30) + 190(last 20) = 900

and althought the answer is 900, but based on the answers, it should be 100+600+200 = 900.

*note: physics formulae cannot be used*

thanks for your help!

ok, lets break it up into the intervals { and PHYSICS FORMULAS CAN BE USED, just they must be DERVIED, i will not derive them as they are standard.}

let u= initial velocity and v= final velocity for each time interval

ok for first 10 seconds, u= 0, a=+2, and t=10

s= ut +(1/2) at^2
= (1/2)(2) (100)
= 100m

for next 30 seconds, the velocity is constant, and equals the velocity at the end of the accelerated interval. v= u +at { this should look familiar, its just like when you had -g as your y acceleration, and integrated to find to y velocity, and your constant of integration was your initial y velocity}

v= 0 + (2)(10) = 20m/s

therefore distance = velocity * time= 20 * 30= 600m

for last 20 seconds, u= 20m/s, a= -1 and t=20

s = ut +(1/2)at^2
= 20(20) +(1/2) (-1) (400)
= 200m

therefore total = 100 + 200 + 600= 900m

PHYSICS FORMULAS CAN BE USED, just they must be derived, the physics formulas are the exact same things you use in mathematics motion questions { look at s = ut +(1/2) at^2, that should remind you of the y co ordinate of a projectile!}. Its important and interesting to see where the physics formulas come from, THEY ARE THE EXACT SAME, they didnt just pop out of no where.

the way you did was right, doing the 1 second at a time thing and adding them up, but its very long and will cause mistakes

 

[pi-ka-chew]

thanks for your help...but at the beginning of the exercise, it does say:

"This course requires that even problems where the acceleration is constant must be solved by integration of the acceleration function. Many readers will know three useful equation for motion (which includes the one that you used). These equations automate the integration process, and so cannot be used in this course"

that's where i was stumped.

but nevertheless, thanks for your assistance

 

[random-1006]

thanks for your help...but at the beginning of the exercise, it does say:

"This course requires that even problems where the acceleration is constant must be solved by integration of the acceleration function. Many readers will know three useful equation for motion (which includes the one that you used). These equations automate the integration process, and so cannot be used in this course"

that's where i was stumped.

but nevertheless, thanks for your assistance

look at the physics formulas, they are just generalisation of the maths formulas, thats how you can easily derive them!, its just like how you would derive the projectile motion ones

 

[pi-ka-chew]

look at the physics formulas, they are just generalisation of the maths formulas, thats how you can easily derive them!, its just like how you would derive the projectile motion ones

yeah i know but it really does say we can't use them. reinforcing the statement at the beginning of the exercise, a question later on says to derive the 3 formulae and specifically quoted that "they are not to be used elsewhere".
i'm really frustrated about this.

 

[jet]

look at the physics formulas, they are just generalisation of the maths formulas, thats how you can easily derive them!, its just like how you would derive the projectile motion ones

It's not in the syllabus so you shouldn't be using them at all. It's better off doing it the way the markers recognise and know, rather than ways they aren't expecting to encounter to give yourself the best chance at getting the marks. They aren't assessing you on your ability to derive formulae they're assessing you on your ability to integrate acceleration functions of time to yield velocity and displacement.

 

[random-1006]

It's not in the syllabus so you shouldn't be using them at all. It's better off doing it the way the markers recognise and know, rather than ways they aren't expecting to encounter to give yourself the best chance at getting the marks. They aren't assessing you on your ability to derive formulae they're assessing you on your ability to integrate acceleration functions of time to yield velocity and displacement.

well they are actually, the derivation is done through the integration

 

[random-1006]

ok, consider a particle undergoing motion in one dimension with constant acceleration "a"

acceleration= a { integrate boths sides with respect to t)
velocity = at +C ( but when t=0, v=u, therefore C= u
velocity= u + at { integrate boths sides with respect to t}
displacement= ut + (1/2)at^2 +c { take initial displacement as 0, therefore c= 0}

therefore s = ut +(1/2)at^2 { now we know from the question that the car does not turn around, so the displacement will represent the distance travelled, then apply that formula in the three different intervals}

done , easy!

there should be NO reason why that should get marks deducted, you have derived/proved the formula you used { just in case check with your own maths teacher, but it shouldnt lose marks}, and its the exact same sort of thing you do in 3 unit projectiles

 

[random-1006]

^ it does matter if you use the formulas WITHOUT DERIVING THEM, that will surely lose marks

see my last post for how to derive

 

[pi-ka-chew]

ah i see. thanks for all your help!