Question

According to recent typical test data, a Ford Focus travels 0.250mi0.250 \mathrm{mi} in 19.9 s19.9 \mathrm{~s}, starting from rest. The same car, when braking from 60.0mph60.0 \mathrm{mph} on dry pavement, stops in 146ft146 \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 0.250mi0.250 \mathrm{mi} of acceleration?
(c) How long does it take the car to stop while braking from 60.0mph60.0 \mathrm{mph} ?

Solution

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We 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:

a=ΔvΔt\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 axa_{x} is given by

ax=vf,xvi,xtfti\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

vx2=v0,x2+2 ax (xx0)\begin{align*} v_{x}^{2} &= v_{0,x}^{2} + 2 ~ a_{x} ~ \left( x - x_{0} \right) \\ \end{align*}

As equation (2.122.12) mentions that the traveled distance by an object moves with constant acceleration equals to

x=x0+v0,x t+12 ax t2\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 mphmph and ft/sft/s is 1 mph=1.46667 ft/s1 \mathrm{~mph} = 1.46667 \mathrm{~ft/s}. Also, the conversion rate between miles and feet is 1 mi=5280 ft1 \mathrm{~mi} = 5280 \mathrm{~ft}.

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