Question

A body start with a velocity of 0.3m/s move in a straight line with a constant acceleration if it's velocity at the end of 5s is 5.5m/s, find 1.the uniform acceleration and 2.distance travelled in 10s

Answer 1

$\large\bold{\underline{\it{\mathfrak{Given:}}}}$

U = 0.3m/s

U = 0.3m/s V = 5.5m/s

U = 0.3m/s V = 5.5m/s t = 5 seconds

$\large\bold{\underline{\it{\mathfrak{To\:find:}}}}$

1. uniform acceleration

$\large\bold{\boxed{\boxed{a\:=\frac{\:v\:-\:u\:}{t}}}}$

$\large\bold{a\:=\frac{\:5.5\:-\:0.3\:}{5}}$

$\large\bold{a\:=\frac{5.2}{5}}$

$\large\bold{\boxed{\boxed{a\:=\:1.04m/s}}}$

2.Distance travelled in 10 seconds

$\large\bold{\boxed{\boxed{S\:=ut\: +\:\frac{1}{2}a{t}^2}}}$

$\large\bold{S\:=0.3\:×\:10\: +\:\frac{1}{2}\:×\:1.04\:×\:{10}^2}$

$\large\bold{S\:=3\: +\:\frac{1}{2}\:×\:1.04\:×\:100}$

$\large\bold{S\:=3\: +\:\frac{1}{2}\:×\:104}$

$\large\bold{S\:=3\: +\:52}$

$\large\bold{\boxed{\boxed{S\:=55m}}}$

Answer 2

$\huge{\underline{\underline{.........Answer.........}}}$

Given:-

u = 0.3 m/s

v = 5.5 m/s

t = 5 seconds

Explanation:-

• Case-1

Acceleration = ?

From the first law of motion.

$v = u + at$

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

$a = \frac{5.5 - 0.3}{5}$

$a = \frac{5.2}{5}$

$a = 1.04$

$\boxed{\boxed{a = 1.04 m/s^2}}$

So,the acceleration produced is 1.04 m/s^2.

• Case-2

Distance travelled = ?.

From second equation of motion.

$s = ut + \frac{1}{2} a {t}^{2}$

$s = 0.3 \times 10 + \frac{1}{2} \times1.04 \times 100$

$s = 3 + \frac{1}{2} \times 104$

$s = 3 + 52$

$s = 55 \: meters$

$\boxed{\boxed{s = 55 meters}}$

So,the distance travelled is 55 meters.