1. ## Mechanics Speed Question

A balloon moves vertically upwards in a straight line so that at time t seconds its acceleration, a metre per second is
a=3t/200

Initally the balloon has a speeed of 0.5m/s
i)find the speed of the balloon when t=30sec
ii)Find how far the balloon travels in the first 30sec.

Any help much appreciated.

2. hi there,

we have the following data

$\displaystyle Acceleration,a = \frac{3t}{200}$
$\displaystyle InitialSpeed, u= 0.5$
$\displaystyle TimeTaken, t = 30$

to find the final velocity we will use the formula
$\displaystyle v = u+at$

substituting values we have the final velocity can be obtained

to find the distance traveled for this first 30 seconds use this formula with the given information:
$\displaystyle s = ut+\frac{1}{2}at^2$

$\displaystyle Acceleration,a = \frac{3t}{200}$
$\displaystyle InitialSpeed, u= 0.5$
$\displaystyle TimeTaken, t = 30$

again by substituting the given value you should be able to get the required distance traveled ^-^ good luck.

3. yes i have tried using those equationns previously but got answers of 13 etc. the answer i know is infact 7.25.help plz!!

4. ou ok i understand why it is not this method, the previous method was for constant acceleration, however in this case it is for varying acceleration:

the right method to use is to consider:
$\displaystyle \frac{d}{dt}v = a$

$\displaystyle \frac{d}{dt}v = \frac{3t}{200}$

$\displaystyle \int \frac{d}{dt}v = \int \frac{3t}{200}dt$

$\displaystyle v = \frac{3t^2}{400} + c \longrightarrow(1)$

when t = 0, u = 0.5, from (1);
$\displaystyle u = \frac{3(0)^2}{400} + c$
$\displaystyle c = 0.5$

so the equation to determine the speed of the balloon after 30 seconds is:
$\displaystyle v = \frac{3t^2}{400} + 0.5$

$\displaystyle v = \frac{3(30)^2}{400} + 0.5$

this will give you the answer of 7.25 ms^(-1)

for the second part we can also observe just like in the case of finding the velocity:

$\displaystyle \frac{d}{dt}S = v$

$\displaystyle \int \frac{d}{dt}S =\int vdt$

$\displaystyle \int \frac{d}{dt}S =\int \frac{3t^2}{400} + 0.5dt$

$\displaystyle S = \frac{3t^3}{1200} + 0.5t + k \longrightarrow(2)$

at t = 0, S = 0, therefore from equation (2):
$\displaystyle S = \frac{3t^3}{1200} + 0.5t + k$

$\displaystyle 0 = \frac{3(0)^3}{1200} + 0.5(0) + k$
$\displaystyle k = 0$

therefore the general equation of the displacement, S is:

$\displaystyle S = \frac{3t^3}{1200} + 0.5t$