Answer to Question #91667 in Mechanics | Relativity for suhan

Question

An object moving at speed of 20ms-1 loses its speed by 3ms-2 . How does it travel before it stops?

Answer 1

We can find the distance traveled by the object before it stops from the kinematic equation:

$$v_f^2 = v_i^2 + 2ad,$$

here, $v_i = 20 m/s$ is the initial velocity of the object, $v_f = 0$ is the final velocity of the object, $a = -3 m/s^2$ is the deceleration of the object and $d$ is the distance traveled by the object before it stops.

Then, from this formula we can calculate the distance traveled by the object before it stops:

$$d = \dfrac{-v_i^2}{2a} = \dfrac{-(20 \dfrac{m}{s})^2}{2 \cdot (-3 \dfrac{m}{s^2})} = 67 \dfrac{m}{s}.$$

Answer:

$d = 67 \dfrac{m}{s}.$