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An object accelerates at the rate of 1.30 m/s^2 over a displacement of 11.8m, reaching a final velocity of 7.00 m/s. For what length of time was it accelerating? Explain the answers.
NOTE: Actually, when I first was about to post this, I finally had gotten the answer I needed. However, I cannot explain why I got these two answers (So now a new question.
a = 1.30 m/s^2 , d = 11.8m , v = 7 m/s , t = ?
d = v2t - 1/2at^2
11.8 = 7t - 0.65t^2
0 = 0.65t^2 - 7t + 11.8
-b [tex]\pm[/tex][tex]\sqrt{b^2-4ac}[/tex]
/2a (The entire equation above)
7 [tex]\pm[/tex][tex]\sqrt{-7^2-4(0.65)(11.8)}[/tex]
/2(0.65) (The entire equation above)
= 7 [tex]\pm[/tex][tex]\sqrt{49-30.68}[/tex]
/1.3 (The entire equation above)
Answer:
(7 + 4.28)/2 = 8.68 s
or
(7 - 4.28)/2 = 2.09 sNow I know the answer but, I can't explain why there could be two answers. Can anyone explain?
NOTE: Actually, when I first was about to post this, I finally had gotten the answer I needed. However, I cannot explain why I got these two answers (So now a new question.
a = 1.30 m/s^2 , d = 11.8m , v = 7 m/s , t = ?
d = v2t - 1/2at^2
11.8 = 7t - 0.65t^2
0 = 0.65t^2 - 7t + 11.8
-b [tex]\pm[/tex][tex]\sqrt{b^2-4ac}[/tex]
/2a (The entire equation above)
7 [tex]\pm[/tex][tex]\sqrt{-7^2-4(0.65)(11.8)}[/tex]
/2(0.65) (The entire equation above)
= 7 [tex]\pm[/tex][tex]\sqrt{49-30.68}[/tex]
/1.3 (The entire equation above)
Answer:
(7 + 4.28)/2 = 8.68 s
or
(7 - 4.28)/2 = 2.09 sNow I know the answer but, I can't explain why there could be two answers. Can anyone explain?