
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.

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.

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

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
$