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I want to compute the displacement of an object at a certain moment in time, given the values for:

  • Initial velocity vStart (m/s)
  • Target velocity vEnd (m/s)
  • Total distance traveled sTotal (meters)
  • Total duration tTotal (seconds)

At any given time between 0 and tTotal seconds I want to compute the displacement at that moment.

For example, a car travels at 50 m/s, and wants to decelerate to 35 m/s over 200 meters, and knows this will take 15 seconds. I want to compute the car's displacement after e.g. 4 seconds from when he starts decelerating, considering the case where the deceleration is constant.

How would I do this?

M4rw
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1 Answers1

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One of the more common kinematic formulas is the following. $$\Delta x = at^2 + v_1t$$ For clarification, $\Delta x$ is displacement, $a$ is acceleration, $t$ is time, and $v_1$ is the initial velocity.

In the specific question you gave as an example, we aren’t initially given the acceleration, so there are two variables. To find acceleration, you can use the following formula. $$(v_2)^2-(v_1)^2 = 2a\Delta x$$ We are given the following information. $$v_1 = 50 \frac{m}{s}$$ $$v_2 = 35 \frac{m}{s}$$ $$\Delta x = 200 m$$ Now we can solve for $a$. $$35^2-50^2 = 2a(200)$$ $$400a = 1225-2500$$ $$a = -\frac{1275}{400}$$ $$a = -3.1875\frac{m}{s^2}$$ Now, back to the original equation. Since we’ve found $a$, the only variable remaining is $\Delta x$. $$\Delta x = at^2+v_1t$$ Here is the information we have (now that we’ve found $a$). $$a = -3.1875 \frac{m}{s^2}$$ $$t = 4 s$$ $$v_1 = 50 \frac{m}{s}$$ Plug in the values. $$\Delta x = (-3.1875)\cdot(4)^2+50(4)$$ $$\boxed{\Delta x = 149 m}$$ Hopefully this isn’t too long of an answer. Just wanted to keep it as clear as possible.

KM101
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  • Thank you for the explanation. In your example the time it takes to decelerate (15s) isn't taken into account though? – M4rw Oct 04 '18 at 19:29
  • No problem, glad to help. In response to your question, no. In fact, since deceleration is constant, $15 s$ isn’t correct in the first place. Remember, $t = \frac{d}{\overline{v}}$. Average velocity, or $\overline{v}$, is the mean of initial and final velocities, so $\overline{v} = \frac{50+35}{2} = 42.5\frac{m}{s}$. Now, $t = \frac{200}{42.5} = \frac{80}{17} \neq 15s$. As I said though, in these types of problems, it wouldn’t matter anyway. – KM101 Oct 04 '18 at 20:08
  • This is because we’ve found acceleration anyway, which occurs with the passing of time. So the time it takes to accelerate/decelerate from $v_1$ to $v_2$ isn’t useful. – KM101 Oct 04 '18 at 20:14