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受欢迎的代数 >

solvefor t,d=vt+(at^2)/2

解答

求解

t, d = v t + \frac{a t ^{2}}{2}

解答

t = \frac{- v + 2 d a + v ^{2}}{a}, t = - \frac{v + 2 d a + v ^{2}}{a}; a \neq = 0

求解步骤

d = v t + \frac{a t ^{2}}{2}

在两边乘以

2

2 d = 2 t v + t^{2} a

交换两边

2 t v + t^{2} a = 2 d

将

2 d

para o lado esquerdo

2 t v + t^{2} a - 2 d = 0

改写成标准形式

a x^{2} + b x + c = 0

a t^{2} + 2 v t - 2 d = 0

使用求根公式求解

t_{1, 2} = \frac{- 2 v \pm ( 2 v ) ^{2} - 4 a ( - 2 d )}{2 a}

化简

(2 v)^{2} - 4 a (- 2 d) : 2 v^{2} + 2 d a

t_{1, 2} = \frac{- 2 v \pm 2 v ^{2} + 2 d a}{2 a}; a \neq = 0

将解分隔开

t_{1} = \frac{- 2 v + 2 v ^{2} + 2 d a}{2 a}, t_{2} = \frac{- 2 v - 2 v ^{2} + 2 d a}{2 a}

t = \frac{- 2 v + 2 v ^{2} + 2 d a}{2 a} : \frac{- v + 2 d a + v ^{2}}{a}

t = \frac{- 2 v - 2 v ^{2} + 2 d a}{2 a} : - \frac{v + 2 d a + v ^{2}}{a}

二次方程组的解是:

t = \frac{- v + 2 d a + v ^{2}}{a}, t = - \frac{v + 2 d a + v ^{2}}{a}; a \neq = 0

作图

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