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# Mechanics M1 Help will take 1 minute Watch

1. 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
2. (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?
3. (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
4. 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
5. Thanks ive got it correct now U=2, much appreciated

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