# A RACE CAR ACCELERATE UNIFORMLY FROM 18.5M/SEC TO 46.1M/SEC IN 2.47SEC DETERMINE THE ACCELERATION OF CAR AND DISTANCE

2
by arisettynitin

2014-07-11T20:57:11+05:30

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U = 18.5m/sec , V = 46.1m/sec , t = 2.47sec Acceleration = (v-u)/t = 46.1-18.5)/2.47= 27.6/2.47= 11.47 m/sec square now distant acne = ut -1/2 at^2 = 18.5 x 2.47 -1/2 x 2.47x 2.47 x 11.47 = 45.695 - 34.98 = 10.706 m answer
2014-07-11T21:08:10+05:30

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Final velocity (v) = 46.1m/sec
Initial velocity (u) = 18.5m/sec
Time (t) = 2.47secs
Acceleration (a) =?

Formula used:
v=u+at
putting the values,
46.1= 18.5+a×2.47
⇒46.1=18.5+2.47a
⇒46.1-18.5=2.47a
⇒2.47a=28.6
⇒a=
⇒a=11.58 m/sec²

﻿﻿distance (S)=?
acceleration (a) = 11.58 m/sec
²
Initial vel (u) =18.5 m/sec
Final vel (v) = 46.1m/sec

Formula used:
S= ﻿﻿ (u×v)×t

S=  (18.5×46.1)×2.47
S=  852.85×2.47
S=  ×2106.5395
S=1053.27 m