Question

a uniformly accelerated body travels 50metres in 5 seconds . if it covers 14metres during 5th second, find out initial velocity and acceleration?

Answer 1

Hey there !

Solution:

$S_n = u + \frac{1}{2} a \: ( 2n - 1 )$

Given that, 14 meters were covered in the 5th second

$\implies 14 = u + \frac{1}{2}a\: \: ( 2 ( 5 ) - 1 )$

=> 14 = u + 9a / 2

=> 14 = ( 2u + 9a ) / 2

=> 28 = 2u + 9a   => Equation 1

Also, Total distance is given as:

s = ut + at² / 2

=> 50 = u ( 5 ) + a ( 5² ) / 2

=> 50 = 5u + 25a / 2

=> 50 = ( 10u + 25a ) / 2

=> 100 = 10u + 25a

Dividing by 5 throughout the equation we get,

=> 20 = 2u + 5a  

=> 2u = 20 - 5a  => Equation 2

Substituting value of 2a from Equation 2 in Equation 1, we get,

=> 28 = ( 20 - 5a ) + 9a

=> 28 = 20 - 5a + 9a

=> 28 - 20 = 4a

=> 8 = 4a

=> a = 8 / 4 = 2 m/s²

From Equation 2,

2u = 20 - 5a

=> 2u = 20 - 5 ( 2 )

=> 2u = 20 - 10

=> 2u = 10

=> u = 10 / 2 = 5 m/s

Hence the Initial Velocity is 5 m/s and the Acceleration is 2 m/s².

Hope my answer helped !