What is its speed after 1.22 s if it accelerates uniformly at −2.14 m/s 2 ? Answer in units of m/s.?

4 Answer(s)

2.14 m/s^2 for 1.22 s is 2.6108 m/s. That's the change in speed, but you didn't give the starting speed.

Given

Time (t) = 1.22 sec

Acceleration (a) = - 2.14 m/sec²

Final Speed (V) = ?

Let, it was starting from rest, then

Initial speed (u) = 0 m/s

Using Kinematic Relation:

V = u + at

V = 0 + (-2.14) × 1.22

V = - 2.61 m/sec

Can't answer without knowing the initial speed. The relevant equation is:

v = u + at

V = u + at
If it was initially at rest then,
V = 0 + (-2.14)(1.22)
V = -2.6108 m/s