Applied Maths - Linear motion
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Applied Maths - Linear motion

[From: ] [author: ] [Date: 12-07-04] [Hit: ]
iii) if the particle is then brought to rest in 2 ms , what is it deceleration?HELP !-30 km/h = 30*1000/3600 = 25/3 m/s,(i) v = u+at, 25/3 = 0 + 2t ,......
A particle starts from rest with uniform acceleration of 2m/s. Calculate the following giving your answers as rational numbers ( fractions )
i) after how many seconds will its speed be 30 KM/hr ... ans = 25/6
ii) how far from its starting point will the particle be when its speed is 60KM/hr ans = 625/9 ms
iii) if the particle is then brought to rest in 2 ms , what is it deceleration? ans = 625/9 m/s^2
I have the first part done but can't get the other two parts
HELP !

-
30 km/h = 30*1000/3600 = 25/3 m/s, 60 km/h = 50/3 m/s
now apply the SUVAT equations

(i) v = u+at, 25/3 = 0 + 2t , t = 25/6 s <-------

(ii) v² = u² + 2as, (50/3)² = 0 + 4s, s = 2500/36 = 625/9 m <------

(iii) v² = u² + 2as , (50/3)^2 = 0 +2a*2, a = -2500/36 = -625/9 m/s²
[ -625/9 m/s² acceleration ≡ 625/9 m/s² deceleration ]
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keywords: Applied,Linear,motion,Maths,Applied Maths - Linear motion
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