1) use the process of subsitution to derive the uniform acceleration equation in which the initial velocity has been omitted.
2) A car travelling along a highway must uniformly reduce its vwlocity to 12 m/s (North) in 3 seconds. if the displacement travelled during that time interval is 58m (north). What's the car's average acceleration?
What's is its initial velocity?
thx
plz list the steps and ans!
thx a lot!!
Physic motion Question (urgent) !!! ----20 pts
2007-09-07 7:49 pm
回答 (2)
2007-09-08 8:13 am
✔ 最佳答案
1) Consider:v = u + at
u = v - at --- ( 1 )
s = ut + 0.5at^2 --- ( 2 )
Put ( 1 ) into ( 2 ),
s = ( v - at )( t ) + 0.5 at^2
s = vt - at^2 + 0.5at^2
s = vt - 0.5at^2
( 2 ) From the above,
58 = ( 12 )( 3 ) - ( 0. 5 )( a )( 3^2 )
4.5a = -22
a = -4.89 m / s^2 ( cor. to 3 s.f. )
By v = u + at,
12 = u + ( - 4.89 )( 3 )
u = 26.7 m / s ( cor. to 3 s.f. )
2007-09-08 23:59:31 補充：
For Q2, after u've found a = -4.89m/s^2, besides using v = u + at to find u, you can hv the following choices:(1) s = ut + 0.5at^2 58 = 3u + 0.5 ( -4.89 )(3^2)u = 26.7m/s(2) v^2 - u^2 = 2as12^2 - u^2 = 2 (-4.89)(58)u = 26.7m/s
2007-09-08 23:59:45 補充：
(3) s = 0.5(u+v)t58 = 0.5(u+12)(3)26.7m/s The same answer will be obtained no matter which method is used.
參考: My Phy Knowleedge
2007-09-08 8:20 am
1. use the two equations:
v = u+a.t and v^2 = u^2 + 2.a.s
substitute u from the first equation into the second, we have
v^2 = (v-at)^2 + 2.a.s
v^2 = v^2 - 2v.a.t + (at)^2 + 2.a.s
2.a.s = 2.v.a.t - (at)^2
s = v.t - (1/2)a.t^2
2. v = 12 m/s, t = 3 s, s = 58 m, a =?
thus, 58 = 12x3 - (1/2).a.(3x3)
solving for a gives a = - 4.89 m/s2
v = u+a.t and v^2 = u^2 + 2.a.s
substitute u from the first equation into the second, we have
v^2 = (v-at)^2 + 2.a.s
v^2 = v^2 - 2v.a.t + (at)^2 + 2.a.s
2.a.s = 2.v.a.t - (at)^2
s = v.t - (1/2)a.t^2
2. v = 12 m/s, t = 3 s, s = 58 m, a =?
thus, 58 = 12x3 - (1/2).a.(3x3)
solving for a gives a = - 4.89 m/s2
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