Respuesta :

Answer:

Acceleration, [tex]a=-6.705\ m/s^2[/tex]

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 :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{0-26.82\ m/s}{4\ s}[/tex]      

[tex]a=-6.705\ m/s^2[/tex]

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

Answer:

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

Explanation:

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

[tex]\vec{v_0}=(-60mph)\hat{j}[/tex]

The acceleration is given by this kinematic equation:

[tex]\vec{a}=\frac{\vec{v_f}-\vec{v_0}}{t}[/tex]

We do the conversion from seconds to hours:

[tex]4s*\frac{1h}{3600s}=1.1*10^{-3}h[/tex]

Finally, replacing the given values:

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