Question

A body moves in a straight line with a uniform acceleration . if the initial velocity of the body is 5m/s and the velocity after 10s in 15m/s find the acceleration of the body and the distance covered by the body in this time interval

Answer 1

Answer:

Acceleration of the body = 1 m/s²

Distance covered by the body = 100 m

Explanation:

Given that,

Initial velocity = u = 5 m/s

Final velocity = v = 15 m/s

Time taken = t = 10s

We know that,

Acceleration = Change in velocity / Time taken

$\boxed {a = \frac{(v - u)}{t} }$

a = (15 - 5) / 10

a = 10 / 10

a = 1 m/s²

From Newton's Laws of motion,

$\boxed {S = ut + \frac{1}{2}at^{2} }$

S = 5×10 + 1/2 × 1 × 10²

S = 50 + 1/2 × 1 × 100

S = 50 + 50

S = 100 m

Answer 2

Answer:

Given :

Initial velocity ,u = 5m/s

Final velocity,v = 15 m/s

Time taken ,t = 10sec

To Find :

•Acceleration

•Distance covered

Solution:

Using first equationof motion :

$\boxed{\green{V=U+AT}}$

Where ,

v is Final velocity

u is initial velocity

t is time taken

and a refers to acceleration produced

$\displaystyle{\implies 15=5+a\times10}$

$\displaystyle{\implies 15 = 5 +10a}$

$\displaystyle{\implies 15-5=10a \implies 10a=15-5}$

$\displaystyle{\implies a=\frac{10}{10}}$

$\displaystyle {\implies a =1 m/{s}^{2}}$

Using second equation of motion :

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

where ,

s refers to distance covered

u refers to initial velocity

t refers to time taken

a refers to acceleration produced

$\displaystyle{\implies s = 5 \times 10 + \frac{1}{2}\times 1 \times {10}^{2}}$

$\displaystyle{\implies s = 50+\frac{1}{2}\times 1\times 100}$

$\displaystyle{\implies s = 50 +\frac{1}{\cancel 2}\times \cancel 100}$

$\displaystyle{\implies s = 50 + 50}$

$\displaystyle{\implies s = 100 m}$