anti404
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hi, so here's a little background on why this may seem such a simple question to answer for others: I've never taken a physics course, or even a math course above the level of high-school precalculus. so here I am as a sophomore in college level intro. physics and have no idea how to solve any of these problems, as we are supposed to have a high-school physics background. alright, enough with that...
1. A young woman named Kathy Kool buys a sports car that can accelerate at the rate of 4.96 m/s2. She decides to test the car by dragging with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.11 s before Kathy. If Stan moves with a constant acceleration of 3.67 m/s2 and Kathy maintains an acceleration of 4.96 m/s2, find the time it takes Kathy to overtake Stan.
if I've read this correctly, we are given a1-4.96 m/s2, a2-3.67 m/s2, xo-0[for both racers], and vo-0[also for both racers]. we are looking for t1, the time at which Kathy's position is equivalent to Stan's position. again, if reading correctly, t1=t2-1.11s. I am using the (1) to reference Kathy and the (2) to reference Stan.
2. x = x_0 + v_0 t + (1/2) a t^2
I would assume, as we are trying to find time at a certain point[the position where both racers have met], given acceleration, that would be the most useful equation.
3. okay, honestly, I don't know what to do.
as we are relating two racers' relative positions and times, I'd assume we are trying to equate something like x= x_01 + v_01(t_01) + (1/2) a_1 (t_1)^2 to x=x_02 + v_02(t_02) + (1/2)a_2 (t_1)^2.
I attempted subtracting the equations to rid myself of variables, and then subbing t_1=t_2-1.11s into the equation, but that was not correct.
okay, well, that is all I know. I know it's not much, and if you are all unable to help, I understand, as that looks to be a jumbled mess.
well, I'll keep trying this myself, and hoping that someone is able to give me any advice, as I am really struggling to keep afloat here.
thanks all, Justin.
1. A young woman named Kathy Kool buys a sports car that can accelerate at the rate of 4.96 m/s2. She decides to test the car by dragging with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.11 s before Kathy. If Stan moves with a constant acceleration of 3.67 m/s2 and Kathy maintains an acceleration of 4.96 m/s2, find the time it takes Kathy to overtake Stan.
if I've read this correctly, we are given a1-4.96 m/s2, a2-3.67 m/s2, xo-0[for both racers], and vo-0[also for both racers]. we are looking for t1, the time at which Kathy's position is equivalent to Stan's position. again, if reading correctly, t1=t2-1.11s. I am using the (1) to reference Kathy and the (2) to reference Stan.
2. x = x_0 + v_0 t + (1/2) a t^2
I would assume, as we are trying to find time at a certain point[the position where both racers have met], given acceleration, that would be the most useful equation.
3. okay, honestly, I don't know what to do.
as we are relating two racers' relative positions and times, I'd assume we are trying to equate something like x= x_01 + v_01(t_01) + (1/2) a_1 (t_1)^2 to x=x_02 + v_02(t_02) + (1/2)a_2 (t_1)^2.
I attempted subtracting the equations to rid myself of variables, and then subbing t_1=t_2-1.11s into the equation, but that was not correct.
okay, well, that is all I know. I know it's not much, and if you are all unable to help, I understand, as that looks to be a jumbled mess.
well, I'll keep trying this myself, and hoping that someone is able to give me any advice, as I am really struggling to keep afloat here.
thanks all, Justin.