Question

A basketball is rolling rightward onto the court with a speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m ​ 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, and slows down with a constant leftward acceleration of magnitude 0.50\,\dfrac{\text m}{\text s^2}0.50 s 2 m ​ 0, point, 50, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction over 14\,\text m14m14, start text, m, end text.

Answer 1

Answer:

Final velocity, v = 1.41 m/s.

Explanation:

It is given that,

Initial velocity of the basketball, u = 4 m/s

Acceleration of the ball, a = -0.5 m/s² (as it slows down)

We need to find the velocity of the basketball after rolling for 14 m. It is a concept of equation of motion in kinematics. Using third equation of motion as :

$v^2=u^2+2ad$

v is final velocity

$v^2=(4)^2+2\times (-0.5)\times 14\\\\v=\sqrt 2\ m/s\\\\v=1.41\ m/s$

So, the final velocity of the basketball is 1.41 m/s.