What is the final speed of the particle?

Question

A particle moving at a velocity of 5.7 m/s in the positive x direction is given an acceleration of 5.3 m/s2 in the positive y direction for 6.7 s. Answer in units of m/s.

Answer 1

In this case, the formula for the velocity is v(t) = v₀ + at, where v is the velocity at time t, v₀ is the initial velocity, a is the acceleration, and t is the time. The two directions are independent of each other, so a force in the y direction will have no effect on the x direction. The numbers given are

 

v_0x = 5.7 m/s

v_0y = 0

a_x = 0

a_y = 5.3 m/s²

t = 6.7 s

 

Thus,

 

v_x = v_0x = 5.7 m/s

v_y = (5.3 m/s²)(6.7 s) = 35.51 m/s

 

So the velocity vector is (5.7,35.51), which means the speed (the magnitude of the vector is)

 

speed = sqrt[5.7² + 35.51²] = 35.96 m/s