Mechanics help needed

[ahmed t]

16)well first we must show that decelerating time is 6 times more than accelerating time
for accelerating part we know that
a=10/t so the accelerating time is t=10/a
as for decceleration we know 0.5a=30/T (notice the capital T)
so T=60/a
soooo deccelerating time T is six times as much as t
now part i)
we know all the area under the graph is equal to 2900 so we split the are into shapes(trapezium rectangle and triangle)
soo
0.5*(20+30)*t +600(rectangle) + 0.5*30*T (we know T=6t so we can replace it)
we solve for t

 

[ahmed t]

some ansers are wrong in the book
im sure of my anser my teacher cheked it

 

[arlery]

hey don't solve 22, I've already done that, instead can you solve 23 c

 

[ahmed t]

back to Q16
ii)now we have to all we do is replace it into the equation
t=10/a

iii)well 40 seconds before it comes to rest
we know the last 40 seconds are in the decceleration part and we can calculate the decceleration part to be 120 seconds
so the time it travels while deccelerating is 120-40 which is 80
we ma use an equation such as
V=u+at
U is 30 A is -0.25 and T is 80
V is there for 10

 

[ahmed t]

23 C)well u find out how much it falls in three seconds which will be 45m
and then minus it from how much it travels in two seconds which is 20
which will be 25
remember they want it during THE third second
and not during THREE seconds

 

[arlery]

Ohh thanks !!!!!!!!

 

[ahmed t]

if u need anything else im here to help

 

[arlery]

If you're already so good at P4 why do u need mechanics notes?

 

[arlery]

Q. 28 . BTW u got a PM !

 

[ahmed t]

different notes may have different approaches to questions

oh are u takin Alevel or As

 

[arlery]

AS

 

[ahmed t]

q28
well time taken goin up is the same as time taken comin down so it will take 1.5 seconds to rise and another 1.5 to fall
now use s=0.5at^2 a here is 10 t is 1.5 ie time taken to fall
that will give u 11.25
ii)use the formula s=ut+0.5at^2
s=11.25
t=1.5
a=-10
find u it will be 15

 

[arlery]

Thanks !! so whenever an object is thrown & caught at the same height, the total time is divided equally?

 

[ahmed t]

yep but it must be caught at the same hieght
this is because the acceleration is always the same goin up and the same goin down

 

[arlery]

In q. 29 first I got the time taken to travel upwards which is 0/7s, then the time for downward motion which is 1.81s. Why do I have to subtract .7s from 1.8.1 to get 1.11 s??

 

[ahmed t]

another mistake in the book
all u must do is add but he must of subtracted
thats why i hate this book

 

[arlery]

True, a lot of answers are wrong.

 

[arlery]

Q 31

 

[ahmed t]

well
distance BC is 420
speed =distance/time
speed is 3
distance is 420
time will be 140s


ii)the time taken for the cyclist to reach C is 140 same as the runner because they meet up at C
so we can get the speed by s=0.5(u+v)t
s=980
t=140
u=0
then v will be 14

 

[arlery]

Forces q. 4