Question

help please!

A particle at rest undergoes an acceleration of 2.5 m/s2 to the right and 4 m/s2 up.
What is its speed after 5.1 s? Answer in units of m/s

Answer 1

@Mertsj can you help me please?

Answer 2

I think you should find the hypotenuse of the right triangle that is the acceleration and then take the integral. Then replace t with 5.1

Answer 3

You may not need integration...all you need is to see that:

So then you have:
\[||a_{av}||=\sqrt{2.5^2+4^2}=\sqrt{22.5}\]
Then you know that:
\[||a_{av}||=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{\Delta t}=\frac{v_f}{\Delta t}\]
I assumed that \(v_i=0m/s\) since we are starting from rest so therefore:
\[v_f=||a_{av}||\Delta t=\sqrt{22.5}\times5.1=5.1\sqrt{22.5}\]