△ X = V 0 T + 1 / 2 a T 2 − 45 m = 10 m / s T − 4.90 m / s 2 T 2

Question

△  X  =      V          0        T  +  1      /    2  a      T          2        −  45  m  =  10  m      /    s  T  −  4.90  m      /        s          2            T          2

umshikepl · Accepted Answer

Since you have two equations, and both have a T and

T^{2}

term, you can rewrite the second equation to have

T^{2}

interms of

T : - 4.90 m / s^{2} T^{2} = - 45 m - 10 m / s T

.
Multiply through by

(- 1) : 4.90 m / s^{2} T^{2} = 45 m + 10 m / s T

.
Divide through by

(4.90 m / s^{2}) : T^{2} = (45 s^{2} / 4.90) + (10 s / 4.90) T

.
Now, plug this equation into the first equation:

△ X = V_{0} T + 1 / 2 a T^{2} = V_{0} T + 1 / 2 a [(45 s^{2} / 4.90) + (10 s / 4.90) T]

.
Multiply through:

△ X = V_{0} T + 45 s^{2} / 9.80 + 10 s T / 9.80

.
Multiply all by

(9.80) : 9.80 △ X = 9.80 V_{0} T + 45 s^{2} + 10 s T

.
Add like terms:

9.80 △ X = T (9.80 V_{0} + 10 s) + 45 s^{2}

. Your solution, then, gives

△ X

in terms of T and

V_{0}

.

triangleX=V_0T+ 1/2aT^2 -45m=10m/s T- 4.90m/s^2 T^2

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