# M1 HelpWatch

#1
I will ad the respoder to my rep list as i cannot rep atm!

A cyclist travels on a straight road with a constance acceleration of 0.6ms^-2 . P and Q are two points on the road. 120 metres apart. Given that teh cyclist increases speed by 6ms^-1 as he travels from P to Q find the speed of P and time taken from P to Q
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12 years ago
#2
u=u
v=u+6
a=0.6
s=120
t=t

v^2 = u^2+2as
u^2 + 12u + 36 + u^2 + 2*0.6*120
12u = 144
u = 12ms^-1

v=u+at
t = (v-u)/a = 6/0.6 = 10s
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12 years ago
#3
Fenchurch you really should be doing your kinematic equations yourself, their really not that difficult all you have to do is write out the variables, just think what have i got?
S=? U=? V=? A=? T=?
it is quite easy to find the different variables from the given text, and then all you have to do is think which equation can i apply with the variables given to get the output i want, if you cannot do these yourself go see your teacher and get them to help you because getting people on TSR to do it will probably not help you all that much.
(THERE RANT OVER)
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