# Velocity and acceleration with displacement [closed]

Find the velocity and the acceleration for a moving object with displacement $= te^{-t}$.

The $e$ is the exponential function.

## closed as off-topic by John Rennie, DavePhD, Qmechanic♦Jun 13 '14 at 13:16

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• To answer this you need to: 1. know the definitions of velocity and acceleration which you can find in any book, 2. be able to compute a simple derivative. – DarioP Jun 13 '14 at 6:54

$$v\equiv \dot{x} = \dot{te^{-t}} = e^{-t} - te^{-t} = -e^{-t}(t-1)$$ $$a \equiv \dot{v} = e^{-t}(-1)+e^{-t}(-1)(1-t)=e^{-t}(t-2)$$

The first thing you need to realize here is that the velocity is the derivative of the displacement with respect to time. This is because the velocity tells you how fast the displacement changes. Therefore, to find the velocity, you need to differentiate the displacement with respect to the time. The displacement is given as a function of time.

$$s=te^{-t}$$

So, how do you differentiate this? In this case, one may apply the product rule!

$$ds/dt=e^{-t}dt/dt+t\frac{d(e^{-t})}{dt}$$ $$v=e^{-t}-te^{-t}$$ $$v=e^{-t}(1-t)$$

Now you have calculated the speed of the object! This is great progress. Let us now calculate the acceleration. The acceleration is how fast the velocity changes, i.e. the derivative of velocity. So in our case,

$$dv/dt=\frac{d(e^{-t})}{dt}-\frac{d(te^{-t})}{dt}$$ $$a=-e^{-t}-e^{-t}(1-t)$$ $$a=e^{-t}(t-2)$$

I hope this answer helped! If it did, you can do me (and the awesome community here) a favor by "accepting this answer" by clicking the checkmark next to the answer.

• I realise that there is already an answer here, but I feel that a bit more explanation can get the point clearer to the person who asked the question. Please tell me if it is inappopriate to answer an answered question, and I will immediately delete it. – Kansas William Jun 13 '14 at 6:58

The important relation that you will use is that if your displacement is $\ell$ then your velocity $v$ will be: $$v\equiv \frac{\mathrm d\ell}{\mathrm dt}.$$ So you need to find the derivative of your displacement with respect to time to get your velocity. So in your example we have: \eqalign{v&=\dfrac{\mathrm d}{\mathrm dt}te^{-t} \\ &= e^{-t}-t{e^{-t}} \text{ by the product rule} \\ &= e^{-t}(1-t). } We also know that acceleration is the derivative of velocity with respect to time. So the acceleration in your case is given by: \eqalign{a&=\dfrac{\mathrm d}{\mathrm dt}v=\dfrac{\mathrm d}{\mathrm dt}e^{-t}(1-t) \\ &=-e^{-t}-(1-t)e^{-t} \text{ by the product rule} \\ &= e^{-t}(t-2) .}