Question

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 1

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.

`v^2_f=v^2_0+2ad`

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.

`55\cdot(mi)/(h)\cdot(1609.34m)/(1mi)\cdot(1h)/(3600\sec)\approx24.6\cdot(m)/(s)`

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

`\begin{gathered} v_f=v_0+at \\ 24.6\cdot(m)/(s)=0+a\cdot11\sec \\ a=(24.6\cdot(m)/(s))/(11s) \\ a\approx2.2\cdot(m)/(s^2) \end{gathered}`

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

Now we are able to find the distance traveled.

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

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