For Q3:
Each small line is 40 Hz and the lowest frequency is the the first line on the graph therefore if you count them until the first frequency line, you will get 440 Hz
Q9 c:
Eq. of a straight line is the form y = mx + c
You know that the graph is of s/t for y-axis and t for x-axis
Therefore your y = s/t and you know that s/t is speed therefore y = v
So eq. will be like:
s = ut + 1/2at2
Dividing by t will give : s/t = u + 1/2at
Substituting for s/t :
v = u + 1/2at ------> y = c + mx
And for part d..m not so sure
Thanx alot. But regrading the Q3. I just found one more way to solve, hopefully more easy and less error prone.
First, take the least frequncy, which in the graph will be 2200 Hz, and divde it by the total number of bar graps, i.e. 5. You get 440.
Which method is correct though? I merely tried to use Logic.
As for Question 9 (d) first calculate the gradient obtained from the graph, i.e 0.04,
Now according to the formula given : s=ut+1/2at^2
As u = 0; Initial Value
S=0+1/2at^2
s/t=1/2at
s/t/t=1/2a
a=s/t/t x 2
a=gradient x 2 [s/t/t = Gradient]
a=0.04x2=0.08 m/s
Now, s=ut+1/2at^2
s-1/2at^2=ut
u=(s-1/2at^2)/t
=s/t-1/2at
=0.86-1/2 x0.08 x2.32
=0.86-0.04 x2.32
=0.7672 m/sec
Hope it is not to confusing.
