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I'm attempting to find how much force a car will experience during cornering. I understand the acceleration acting on the car if it were travelling in a circular path would be $a = v^2/r$.

However, say a car was travelling on a path given by the function $f(x) = x^2$, how would one find the acceleration acting on the car, given the tangential speed is kept constant.

Furthermore, if the brake/pedal is applied, changing the tangential speed, would the net acceleration simply be the vector sum of the acceleration due to changing direction and the acceleration due to change in speed?

Edit: after having a longer think about it, I came up with the following solution, please check if it is correct:

$f(x)=x^2 \rightarrow f'(x)=2x$

The gradient was represented as a unit vector by dividing the gradient by the vector's magnitude:

$\hat v_1=\frac{\begin{pmatrix}2x \\ 1\end{pmatrix}}{\left|\begin{pmatrix}2x \\ 1\end{pmatrix}\right|}$

After the car has travelled $\Delta x$ meters horizontally:

$f'(x) = 2\Delta x$

$\hat v_2=\frac{\begin{pmatrix}2\Delta x \\ 1\end{pmatrix}}{\left|\begin{pmatrix}2\Delta x \\ 1\end{pmatrix}\right|}$

The acceleration is given by the change in velocity divided by the change in time:

$\frac{dv}{dt} = \lim_{\Delta t \rightarrow 0}\frac{\hat v_2-\hat v_1}{\Delta t}$

However, I do not know how to continue from here. Would I have to know the tangential speed to relate $x$ to $\Delta t$ to evaluate the limit?

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closed as off-topic by Aaron Stevens, Jon Custer, rob Jul 16 at 3:54

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  • $\begingroup$ The information is not sufficient enough to answer the question. Please state the problem exactly. $\endgroup$ – Vish Jul 14 at 14:59
  • $\begingroup$ you look for the acceleration if you apply brake force , or this is kinematic problem? $\endgroup$ – Eli Jul 14 at 16:54
  • $\begingroup$ My initial question was the acceleration when a car travels in a path modelled by an arbitrary function without change in tangential speed. My extension was the acceleration when brake force was applied @Eli $\endgroup$ – George Tian Jul 15 at 2:12
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You are right: the net acceleration is the vector sum of the centripetal and tangential acceleration. Of course, in the case you described, the centripetal acceleration will change continuously because of the tangential braking acceleration.

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