Question

A runner slows down from a 9.50 m/s at a rate of 2.30 m/s^2 . (a) How far does she travel in the next 6.00 s? (b) What is her final velocity? (c) Evaluate the result. Does it make sense?

Answer 1

Answer:

(a) d = 15.6 m

(b) v' = - 4.3 m/s

Given:

Initial velocity, v = 9.50 m/s

Acceleration, a = $- 2.30 m/s^{2]$ or deceleration = 2.30$m/s^{2}$

Solution:

(a) For the calculation of the distance covered, we use eqn (2) of motion:

$d = vt + \frac{1}{2}at^{2}$

where

d = distance covered in time t

t = 6 s (Given)

Now,

$d = 9.50\times 6 - \frac{1}{2}\times 2.30times 6^{2}$

d = 15.6 m

(b) For the calculation of her final velocity, we use eqn 1 of motion:

v' = v + at

v' = 9.50 + (- 2.30)(6) = - 4.3 m/s

(c) Since, the final velocity after the body slows down comes out to be negative and the distance covered and displacement of the body are different, it does not make sense.