non uniform acceleration

**ally79** · February 1st 2009, 04:42 AM

I have the following question,

A car starts from rest and moves in a straight line in the manner descibed below:

Period 1 0 to 4s the acceleration is non uniform i.e after 1s the aceleration is 1m/s after 2 secs the aceleration is 2m/s

period 2 t=4-6 the velocity is constant

period 3 from t=6-10 the car brakes uniformly to rest

calculate the total distance the car travelled after 10 secs

I can work out period 2 and 3 but theres nothing in my book about non uniform acceleration.

any help you can give would be appreciatted

**skeeter** · February 1st 2009, 06:24 AM

Period 1 0 to 4s the acceleration is non uniform i.e after 1s the aceleration is 1m/s after 2 secs the aceleration is 2m/s

no other info about about the interval t = 0 to t = 4 other than v(0) = 0 and the above?

there are many models for non-uniform acceleration that one can come up with to fit the given data ...

here's a simple one

$a(t) = t \,\,, \,\, 0 \leq t < 4$

I'm sure that someone else can come up with other models for acceleration that will fit.

**ally79** · February 1st 2009, 06:42 AM

Thats all the information given in the question, forgive me, but i cant understand how to use what you given me

**skeeter** · February 1st 2009, 07:08 AM

in the interval, $0 \leq t < 4$ , if $a(t) = t$ and $v(0) = 0$ , then you can come up with a velocity function for that time interval.

$v(t) = \int t \, dt$

$v(t) = \frac{t^2}{2} + C$

since $v(0) = 0$ ...

$v(t) = \frac{t^2}{2} \, \, , \, \, 0 \leq t < 4$

on the interval, $4 \leq t < 6$ ... (velocity is constant)

$v(t) = 8$

on the interval, $6 \leq t \leq 10$

$v(t) = 20 - 2t$

total distance traveled ...

$d = \int_0^4 \frac{t^2}{2} \, dt + \int_4^6 8 \, dt + \int_6^{10} 20-2t \, dt$

**ally79** · February 1st 2009, 07:15 AM

got it now! thanks

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