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Homework Statement
An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 140 km/h. Precisely 1.00s after the speeder passes, the policeman steps on the accelerator; if the police car's acceleration is 2.00m/s2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?
v0p = 95 km/h = 26.389 m/s
v0s = 140 km/h = 38.889 m/s
ap = 2.00 m/s
t1 = 1.00 s
vs = v0s
x = x0 + v0t + 0.5 * a * t2
Let t2 = the time it takes them to meet after the initial 1.00s.
After 1.00s, the xp = 26.389m and xs = 38.889m. Taking these two quantities as our initial positions, x0p and x0s, we have:
xp = x0p + v0pt2 + 0.5 * ap * t22
xs = x0s + v0st2
Setting these two equal, I get a quadratic equation and find t2 = 13.43 s or -0.9307 s. The total time would then be 14.43s...
But wait. Shouldn't the negative answer for t2 be -1.00s?? Because 1.00s in the past, the cars met... So I must be doing something wrong...
An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 140 km/h. Precisely 1.00s after the speeder passes, the policeman steps on the accelerator; if the police car's acceleration is 2.00m/s2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?
v0p = 95 km/h = 26.389 m/s
v0s = 140 km/h = 38.889 m/s
ap = 2.00 m/s
t1 = 1.00 s
vs = v0s
Homework Equations
x = x0 + v0t + 0.5 * a * t2
The Attempt at a Solution
Let t2 = the time it takes them to meet after the initial 1.00s.
After 1.00s, the xp = 26.389m and xs = 38.889m. Taking these two quantities as our initial positions, x0p and x0s, we have:
xp = x0p + v0pt2 + 0.5 * ap * t22
xs = x0s + v0st2
Setting these two equal, I get a quadratic equation and find t2 = 13.43 s or -0.9307 s. The total time would then be 14.43s...
But wait. Shouldn't the negative answer for t2 be -1.00s?? Because 1.00s in the past, the cars met... So I must be doing something wrong...