183k views
4 votes
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?

User Lidkxx
by
8.2k points

1 Answer

5 votes

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.

User Jie
by
9.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories