Question

A body is in motion along a straight line as it crosses a fixed point a stop watch is started the body travels a distance of 1.80m in the first 3 second and 2.20 m in next 5 second what will be the velocity at the end of 9 second

Answer 1

Answer:

The velocity at the end of 9 second is 0.3 m/s.

Explanation:

Given that,

Distance d = 1.80 m in the first 3 sec

Distance d '= 2.20 m in the next 5 sec

Let u be the initial velocity and a be the acceleration.

Using equation of motion

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

For t = 3 sec

$1.80=3u+\dfrac{1}{2}\times a\times(3)^2$

$1.80=3u+4.5a$....(I)

For t = (3+5),

$1.80+2.20=(3+5)u+\dfrac{1}{2}\times a\times(3+5)^2$

$4.00=(3+5)u+32a$

$4.00=8u+32a$....(II)

From equation (I) and (II)

$2.40=-60a$

$a = -0.04\ m/s^2$

Put the value of a in equation (I)

$1.80=3u-4.5\times0.04$

$u=\dfrac{1.80+4.5\times0.04}{3}$

$u = 0.66\ m/s$

The velocity at the end of 9 second

$v = u+at$

$v =0.66-0.04\times9$

$v = 0.3\ m/s$

Hence, The velocity at the end of 9 second is 0.3 m/s.

Answer 2

Explanation:

$\huge \sf0.3 \: \: m {s}^{ - 1}$