Verified Answer :

Answer:

Acceleration, \(a=-6.705\ m/s^2\)

Explanation:

It is given that,

Final velocity of the car, v = 0

Initial velocity of the car, u = 60 mph = 26.82 m/s

It comes to rest in 4 s

We need to find its acceleration. The rate of change of velocity is called its acceleration. Mathematically, it is given by :

\(a=\dfrac{v-u}{t}\)

\(a=\dfrac{0-26.82\ m/s}{4\ s}\)      

\(a=-6.705\ m/s^2\)

So, the car is decelerating with a magnitude of 6.705 m/s^2. hence, this is the required solution.

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Answer:

\(\vec{a}=(5.45*10^4\frac{mi}{h^2})\hat{j}\)

Explanation:

Using the y-axis as a reference system, for which the north is positive and the south is negative, we have:

\(\vec{v_0}=(-60mph)\hat{j}\)

The acceleration is given by this kinematic equation:

\(\vec{a}=\frac{\vec{v_f}-\vec{v_0}}{t}\)

We do the conversion from seconds to hours:

\(4s*\frac{1h}{3600s}=1.1*10^{-3}h\)

Finally, replacing the given values:

\(\vec{a}=\frac{0-(-60mph)\hat{j}}{1.1*10^{-3}h}\\\vec{a}=(5.45*10^4\frac{mi}{h^2})\hat{j}\\\)

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