## Related questions with answers

According to recent typical test data, a Ford Focus travels $0.250 \mathrm{mi}$ in $19.9 \mathrm{~s}$, starting from rest. The same car, when braking from $60.0 \mathrm{mph}$ on dry pavement, stops in $146 \mathrm{ft}$. Assume constant acceleration in each part of its motion, but not necessarily the same acceleration when slowing down as when speeding up.

(a) Find this car's acceleration while braking and while speeding up.

(b) If its acceleration is constant while speeding up, how fast (in mph) will the car be traveling after $0.250 \mathrm{mi}$ of acceleration?

(c) How long does it take the car to stop while braking from $60.0 \mathrm{mph}$ ?

Solution

VerifiedWe will consider that in this physical situation, the instantaneous acceleration of any object is given by the changing in velocity with respect to the time interval:

$\begin{align*} a &= \dfrac{ \Delta v }{ \Delta t } \end{align*}$

Since the simplest accelerated motion is a straight line motion with constant acceleration. Then the velocity changes at the same rate throughout the motion. Therefore, the constant or the instantaneous acceleration $a_{x}$ is given by

$\begin{align*} a_{x} &= \dfrac{ v_{f,x} - v_{i,x} }{ t_{f} - t_{i} } \\ \end{align*}$

So, the velocity as a function of position for an object moving with constant acceleration is equal to

$\begin{align*} v_{x}^{2} &= v_{0,x}^{2} + 2 ~ a_{x} ~ \left( x - x_{0} \right) \\ \end{align*}$

As equation ($2.12$) mentions that the traveled distance by an object moves with constant acceleration equals to

$\begin{align*} x &= x_{0} + v_{0,x} ~ t + \dfrac{1}{2} ~ a_{x} ~ t^{2} \\ \end{align*}$

As we know the conversion rate between $mph$ and $ft/s$ is $1 \mathrm{~mph} = 1.46667 \mathrm{~ft/s}$. Also, the conversion rate between miles and feet is $1 \mathrm{~mi} = 5280 \mathrm{~ft}$.

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