1. particle moving 2

a particle travelling in a straight line is subject to a retardation of initially 10 m/s^2 and increases uniformly with time to 13 m/s^2 3 seconds after the start. the initial speed of the particle is 100 m/s. find an expression for the acceleration of the particle in terms of t, the time taken and hence show that the time taken for the particle to come to rest is 10( √3 -1)s

maybe you have to use UVAST equations

2. Re: particle moving 2

Originally Posted by edwardkiely
a particle travelling in a straight line is subject to a retardation of initially 10 m/s^2 and increases uniformly with time to 13 m/s^2 3 seconds after the start.
Taking t= 0 at the start, this says that the "retardation" (deceleration) is 10+ t so the "acceleration" is -10- t.

the initial speed of the particle is 100 m/s. find an expression for the acceleration of the particle in terms of t,

the time taken and hence show that the time taken for the particle to come to rest is 10( √3 -1)s

maybe you have to use UVAST equations
The standard form for that requires constant acceleration which is not the case here. The acceleration is the derivative of the speed so the speed is the anti-dervativie or integral of the acceleration. What is the anti-derivative of -10+ 3t? That will of course involve an "undetermined constant" which will be the initial speed, 100 m/s. Once you have determined the speed function, set it equal to 0 and solve for time.

3. Re: particle moving 2

the integral of -10+3t is -10t/2 +(3t^2)/2

anyway so what I have tried is dv/dt =10+t

dv=(10+t)dt

i then integrate this

v=10t+.5t^2 +c

at t=0 v=100 therefore c=100

so I let v=0

I then get the equation t^2+20t+200 which does not yield the correct answer.

4. Re: particle moving 2

Originally Posted by edwardkiely
the integral of -10+3t is -10t/2 +(3t^2)/2

anyway so what I have tried is dv/dt =10+t

dv=(10+t)dt

i then integrate this

v=10t+.5t^2 +c

at t=0 v=100 therefore c=100

so I let v=0

I then get the equation t^2+20t+200 which does not yield the correct answer.
$a=-(10+t)$ (like Hall's wrote)

try again

5. Re: particle moving 2

Originally Posted by edwardkiely
the integral of -10+3t
That's not the correct acceleration. It should be, as I suggested, -10- 3t= -(10+ 3t)

is -10t/2 +(3t^2)/2
No, it's not. The integral of -10+ 3t is -10t+ (3t^2)/2. There is no "1/2" in the first term. But as I said, that is not correct.

anyway so what I have tried is dv/dt =10+t

dv=(10+t)dt
And this not what you said above! What you had above would be dv= (-10+ t)dt.
What you should have is dv= (-10- t)dt

[quote[i then integrate this

v=10t+.5t^2 +c[/quote]
Okay, you don't have the incorrect "10/2" here. But you should have
v= -10t- .5t^2+ c.

at t=0 v=100 therefore c=100
Even with the minuses, yes, -10(0)- 0,5(0)^2+ c= c= 1000.

so I let v=0

I then get the equation t^2+20t+200 which does not yield the correct answer.[/QUOTE]
With v= -10t- .5t^2+ 1000= 0, we can multiply by -1 and t^2+ 20t- 2000= 0. Note "2000", not "200". You have been awfully careless. There are several places where you write one thing on one line then have it changed on the next!

6. Re: particle moving 2

Originally Posted by HallsofIvy
Even with the minuses, yes, -10(0)- 0,5(0)^2+ c= c= 1000.
I thought the OP said that at t=0, v=100, but you put 1000? I'm confused.

7. Re: particle moving 2

thanks, its working out