Question

object travelling at 24 m/s and decreases it's velocity to 20 m/s in 2 seconds find retardation and distance travelled​

Answer 1

Answer:

The retardation is

$2m {s}^{ - 2}$

and the distance travelled is

$44m$

Explanation:

$u = 24 \: m {s}^{ - 1}$

$v = 20m {s}^{ - 1}$

Acceleration is defined as the rate of change of velocity.

$a = \frac{v - u}{t}$

Using above formula:

$a = - 2m {s}^{ - 2}$

Using third equation of motion.

${v}^{2} - {u}^{2} = 2as$

$s = 44m$

Answer 2

Answer:

a = - 2 m / sec²

s = 44 m

Explanation:

Given :

Initial velocity u = 24 m / sec

Final velocity v = 20 m / sec

Time t = 2 sec

We are asked to find retardation.

From equation of motion we have :

v = u + a t

Putting values here we get :

20 = 24 + 2 a

2 a = 20 - 24

2 a = - 4

a = - 4 / 2

a = - 2 m / sec²

Therefore , retardation of object is 2 m / sec².

From third equation we have :

v² = u² + 2 a s

Putting values here we get :

20² = 24² - 2 × 2 s

400 = 576 - 4 s

4 s = 176

s = 44 m

Hence , distance covered by object is 44 m .