# Mechanics Speed Question

• May 2nd 2008, 12:59 PM
helpishere
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.
• May 2nd 2008, 01:31 PM
hi there,

we have the following data

$
Acceleration,a = \frac{3t}{200}
$

$
InitialSpeed, u= 0.5
$

$
TimeTaken, t = 30
$

to find the final velocity we will use the formula
$
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:
$
s = ut+\frac{1}{2}at^2
$

$
Acceleration,a = \frac{3t}{200}
$

$
InitialSpeed, u= 0.5
$

$
TimeTaken, t = 30
$

again by substituting the given value you should be able to get the required distance traveled ^-^ good luck.
• May 2nd 2008, 01:38 PM
helpishere
yes i have tried using those equationns previously but got answers of 13 etc. the answer i know is infact 7.25.help plz!!
• May 2nd 2008, 02:00 PM
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:
$
\frac{d}{dt}v = a
$

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

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

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

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

$
c = 0.5
$

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

$
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:

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

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

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

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

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

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

$
k = 0
$

therefore the general equation of the displacement, S is:

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