Year 12s: what will be the first way you research your future options? >>

Mechanics M1 Help will take 1 minute

I need help on the last part of this question please, i think its simultaneous eq's but i dont know where to begin, thanks

Reply 1

ghostwalker

Original post by Sniperdon227

I need help on the last part of this question please, i think its simultaneous eq's but i dont know where to begin, thanks

Doesn't need simultaneous equations.

Draw on the extra bit, constant retardation for a further 4 seconds.

Your final right-angled triangle created, has an area of 20, base of 4, and height of 5U. Can you finish?

Reply 2

CasioGamer98

Original post by Sniperdon227

I need help on the last part of this question please, i think its simultaneous eq's but i dont know where to begin, thanks

S 20 M
U 5U m/s
V 0 m/s
A a
T 4

S=0.5(u+v)t => 20=0.5(5U)4 => U=2

might not even be right, who knows

Reply 3

TSRPAV

Done the first part but will show you a method for the last bit since you asked just for that,

so at t=5s v=5u m/s, we are told it comes to rest so let 5u=initial velocity so u=5u m/s, it comes to rest 20 meters away so s=20m and since at rest v=0 m/s and this happens 4 seconds later so t=4s

so s=20m u=5u m/s v=0 m/s t=4s

use s=((u+v)/2)t so 20=((5u+0)/2)*4 so 20=10u so u= 2m/s