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Lexadis
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Solved~Rearranging numbers changes answers?
A car starts from rest and travels with a constant acceleration of [itex]3ms^{-2}[/itex], while a bike which is at a distance of 100m away from the car starts with an initial velocity of [itex]5ms^{-2} [/itex] travels with a constant acceleration of [itex]2ms^{-2}[/itex]. The displacement traveled by the bike before being overtaken is [itex]x[/itex]. Using equations of motion, find,
(i) the time taken for the car to overtake the bike.
(ii) the distance traveled by the bike (x)
(iii) the distance traveled by the car.
[itex]s= ut + \frac{1}{2} at^{2}[/itex]
So I did using the theory that the displacement of the car will be equal to displacement of the bike +100m.
So here it goes:
[itex]ut+ \frac{1}{2}at^{2}+100 = ut+ \frac{1}{2}at^{2} [/itex]
[itex]\frac{1}{2}.3.t^{2} + 100 = 5t + \frac{1}{2}.2.t^{2}[/itex]
[itex]\frac{3}{2}t^{2}+100=5t+t^{2}[/itex]
[itex]100=5t-\frac{1}{2}t^{2}[/itex]
[itex]\frac{1}{2}t^{2}-5t+100=0[/itex]
[itex]t^{2}-10t+200=0[/itex]
It is after this the problem started. I got 2 different answers. Here it goes:
[itex]t^{2}-10t+200=0[/itex]
--> Here I arranged it as -20t + 10t:
[itex]t^{2} - 20t + 10t + 200 = 0[/itex]
[itex] t(t-20)+10(t+20)=0[/itex]
[itex] (t+10)(t-20)(t+20) = 0 [/itex]
[itex] t+10 = 0 / t^{2}-20^{2} = 0[/itex]
[itex]t=-10 / t=20 s / t=-20s[/itex]
-->Here I arranged it as +10, -20, and got different answers >_>
[itex] t^{2} + 10t - 20t + 200 = 0[/itex]
[itex] t(t+10) - 20(t-10) = 0[/itex]
[itex](t-20) (t^{2}-10^{2}) = 0[/itex]
[itex]t=20s / t=10 / t=-10[/itex]
The answer is supposed to be 20. So how come I also got t=10s in my second arrangement? Is there some mistake I can't identify? Thank you very much in advance c:
Homework Statement
A car starts from rest and travels with a constant acceleration of [itex]3ms^{-2}[/itex], while a bike which is at a distance of 100m away from the car starts with an initial velocity of [itex]5ms^{-2} [/itex] travels with a constant acceleration of [itex]2ms^{-2}[/itex]. The displacement traveled by the bike before being overtaken is [itex]x[/itex]. Using equations of motion, find,
(i) the time taken for the car to overtake the bike.
(ii) the distance traveled by the bike (x)
(iii) the distance traveled by the car.
Homework Equations
[itex]s= ut + \frac{1}{2} at^{2}[/itex]
The Attempt at a Solution
So I did using the theory that the displacement of the car will be equal to displacement of the bike +100m.
So here it goes:
[itex]ut+ \frac{1}{2}at^{2}+100 = ut+ \frac{1}{2}at^{2} [/itex]
[itex]\frac{1}{2}.3.t^{2} + 100 = 5t + \frac{1}{2}.2.t^{2}[/itex]
[itex]\frac{3}{2}t^{2}+100=5t+t^{2}[/itex]
[itex]100=5t-\frac{1}{2}t^{2}[/itex]
[itex]\frac{1}{2}t^{2}-5t+100=0[/itex]
[itex]t^{2}-10t+200=0[/itex]
It is after this the problem started. I got 2 different answers. Here it goes:
[itex]t^{2}-10t+200=0[/itex]
--> Here I arranged it as -20t + 10t:
[itex]t^{2} - 20t + 10t + 200 = 0[/itex]
[itex] t(t-20)+10(t+20)=0[/itex]
[itex] (t+10)(t-20)(t+20) = 0 [/itex]
[itex] t+10 = 0 / t^{2}-20^{2} = 0[/itex]
[itex]t=-10 / t=20 s / t=-20s[/itex]
-->Here I arranged it as +10, -20, and got different answers >_>
[itex] t^{2} + 10t - 20t + 200 = 0[/itex]
[itex] t(t+10) - 20(t-10) = 0[/itex]
[itex](t-20) (t^{2}-10^{2}) = 0[/itex]
[itex]t=20s / t=10 / t=-10[/itex]
The answer is supposed to be 20. So how come I also got t=10s in my second arrangement? Is there some mistake I can't identify? Thank you very much in advance c:
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