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A car accelerates uniformly from rest to a speed of 55.0 mi/h in 11.0 s.(a) Find the distance the car travels during this time. m(b) Find the constant acceleration of the car.m/s²

Answer :

We know that

• The initial speed is zero (it starts from rest).

,

• The final speed is 55 mi/h.

,

• The time elapsed is 11 seconds.

To find the distance traveled, let's use the following formula.

[tex]v^2_f=v^2_0+2ad[/tex]

But first, we have to transform 55 mi/h into m/s. We know that 1 mile is equivalent to 1609.34 meters, and 1 hour is equivalent to 3600 seconds.

[tex]55\cdot\frac{mi}{h}\cdot\frac{1609.34m}{1mi}\cdot\frac{1h}{3600\sec}\approx24.6\cdot\frac{m}{s}[/tex]

Then, we find the acceleration because we need it to find the distance traveled.

[tex]\begin{gathered} v_f=v_0+at \\ 24.6\cdot\frac{m}{s}=0+a\cdot11\sec \\ a=\frac{24.6\cdot\frac{m}{s}}{11s} \\ a\approx2.2\cdot\frac{m}{s^2} \end{gathered}[/tex]

(b) The constant acceleration is 2.2 m/s^2.

Now we are able to find the distance traveled.

[tex]\begin{gathered} v^2_f=v^2_0+2ad\to(24.6\cdot\frac{m}{s})^2=0^2+2(2.2\cdot\frac{m}{s^2})d \\ 605.16\cdot\frac{m^2}{s^2}=4.4\cdot\frac{m}{s^2}\cdot d \\ d=\frac{605.16\cdot\frac{m^2}{s^2}}{4.4\cdot\frac{m}{s^2}} \\ d\approx137.5m \end{gathered}[/tex]

(a) The distance the car travels during this time is 137.5 meters.

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