Question

A particle of 4kg mass travelling at 20 m/s
due east is acted upon by force 16 newton
west from t = 0 to t = 10 sec distance
travelled in 0 to 6 sec is :-
(A) 22m
(B) 28 m
(C) 48 m
(D) 52 m​

Answer 1

Explanation:

First by acceleration which is

$a = \frac{f}{m} = \frac{16}{4} = 4 \frac{m}{ {s}^{2} }$

Now distance

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 20\times 6 - \frac{1}{2} \times 4 \times 36 = 48m$

**Negative sign because acceleration is against motion

Answer 2

The distance covered by the particle in 0 to 6 seconds is 48 meters.

Explanation:

It is given that,

Mass of the particle, m = 4 kg

Speed of the particle, u = 20 m/s (due east)

Force, F = -16 N (due west)

We need to find the distance covered by the particle in 0 to 6 seconds. If a is the acceleration of the particle. It is given by using second law of motion as :

F = ma

$a=\dfrac{F}{m}$

$a=\dfrac{-16\ N}{4\ kg}$

$a=-4\ m/s^2$

Let d is the distance covered by the particle in 0 to 6 seconds. It is given by second equation of motion as :

$d=ut+\dfrac{1}{2}at^2$

$d=20\times 6+\dfrac{1}{2}\times (-4)\times (6)^2$

d = 48 meters

So, the distance covered by the particle in 0 to 6 seconds is 48 meters. Hence, this is the required solution.

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