Mechanics 1 HELP!!

Can anyone help im stuck on q. 7 January 03..OCR btw

7. A particle starts from rest at the point O and travels in a straight line until it comes to rest at the point A. Its Velocity vms-1, t seconds after leaving O is given by: v= 0.3t - 0.003t2<----(squared!)

(i)Find the distance OA

(ii) Show that for the motion from O to A, the maximum speed of the particle is 1.5 times its average speed.

i can do sum of it but then i kinda get lost half way thru.....lol, anyhelp wud be appreciated, esp since i suck at mechanics!

Thanx..

Reply 1

Revenged

15

zaaks786

Can anyone help im stuck on q. 7 January 03..OCR btw

7. A particle starts from rest at the point O and travels in a straight line until it comes to rest at the point A. Its Velocity vms-1, t seconds after leaving O is given by: v= 0.3t - 0.003t2<----(squared!)

(i)Find the distance OA

(ii) Show that for the motion from O to A, the maximum speed of the particle is 1.5 times its average speed.

i can do sum of it but then i kinda get lost half way thru.....lol, anyhelp wud be appreciated, esp since i suck at mechanics!

Thanx..

ok... this is what i'm doing for m2 though!!

i) when particle is at rest:

0 = t(0.3 - 0.003t)

Therefore, it is at rest

t = 0

and

0.3 - 0.003t = 0s
t = 100s

Next you must intergrate v= 0.3t - 0.003t² using t = 0s and t = 100s the limits

s = [0.3 x 100² / 2 - 0.003 x 100³ / 3] - 0
s = 1500 - 1000
s = 500m

ok... in order to find the maximum speed you need to differenciate

max speed occurs when dx/dt = 0

v = 0.3t - 0.003t2
0 = 0.3 - 0.006t
t = 0.3 / 0.006
t = 50s

v = 0.3(50) - 0.003(50)²
v = 7.5 m/s

Average speed = s / t
= 500 / 100
= 5 m/s

Therefore,

7.5 / 5 = 1.5

I agree, that's a very hard question for m1! :biggrin:

I get the same.

i wrote a plus instead of a minus first though but managed to correct it before you checked it... took me ages to figure out that the problem was just 1 sign though!