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We were given a couple formulas, but the one that immediately stood out to me was the Vfinal = Vinitial + at

so we know the patrol will constantly accelerate "until he pulls next to the speeding car" so Vfinal = 30m/s and Vinitial = 0(cop at rest) and the acceleration is a constant 3m/s^2

so 30 = 3t, t = 10. However when I continue through this lecture, it turns out t = 20. he uses the formula x = xo + v0t + 1/2at^2 and somehow gets a different time than me. I don't understand how it's possible that we are both solving for time but getting different results.

Is it safe to assume that when xo = 0 and v0 = 0 that

sqrt(2x/a) = Vfinal / a

since we cancel out terms and they both equal t?

Pookie
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3 Answers3

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If the patrol car goes after the other starting from rest, then of course that, when it catches it, it must be going faster than the other car. Therefore, the final speed cannot possibly be $30$ m/s.

José Carlos Santos
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Nowhere does it say that the final speed of the cop is 30 m/s

If it was the cop will never reach the person, in order to reach them , they have to travel faster.

We don't know the final speed of the cop.

So the time is 20 seconds.

Neo
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Your mistake was that you tried finding the time it takes for the cop to achieve the same velocity ($30 \frac{m}{s}$). That’s not right for obvious reasons. The speeding car was already traveling at $30 \frac{m}{s}$ by the time the police car was stationary. Instead, you have to find the time it takes for the distance covered to become equal.

Car $1$: $\overline v = 30 \frac{m}{s} \implies \Delta x_1 = \overline v \cdot \Delta t$
Car $2$: $v_1 = 0 \frac{m}{s}; \overline a = 3 \frac{m}{s^2} \implies \Delta x_2 = \frac{1}{2}\overline a\Delta t^2+v_1\Delta t$

Keep in mind $\Delta x$ represents displacement, $\Delta t$ represents time, $v_1$ represents initial velocity, $\overline v$ represents average/uniform velocity, and $\overline a$ represents average/uniform acceleration.

We need to find the $\Delta t$ for which $\Delta x_1 = \Delta x_2$.

$$\Delta x_1 = \Delta x_2$$ $$\overline v\cdot\Delta t= \frac{1}{2}\overline a\Delta t^2+v_1\Delta t$$ $$30\Delta t = \frac{1}{2}3\Delta t^2+0(\Delta t)$$ $$30\Delta t = \frac{1}{2}3\Delta t^2$$ $$30 = 1.5\Delta t$$ $$\boxed{\Delta t = \frac{30}{1.5} = 20s}$$

KM101
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