Question

Carter is jogging with a velocity of 3.8 m/s when he accelerates at 3.7 m/s² for 3 s. How

fast is Carter running now?

Answer 1

Uniformly Varied Rectilinear Motion.

When we start to solve a physics exercise, we get the exercise data:

Initial Speed = 3.8m/s

Acceleration (a)= 3.7 m/s²

Time (t) = 3 sec.

He tells us that Carter treats at a speed of 3.8 m/s, this is the speed, since he asks us, at what speed does he treat now. Then we must calculate the final speed.

The formula to calculate the final velocity is:

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{V_(f)=V_(o)+a*t} \end{gathered}$} }`

We substitute the data in the formula used and solve:

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{V_(f)=3.8 \ (m)/(s)+3.7 \ \frac{m}{\\ot{s^(2)} }*3 \ s } \end{gathered}$} }`

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{V_(f)=3.8 \ (m)/(s)+11.1 \ (m)/(s) } \end{gathered}$} }`

`\boxed{\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{V_(f)=14.9 \ m/s} \end{gathered}$} }}`

Carter now runs a new speed of 14.9 m/s.