Question

a body moving with uniform acceleration travells 50m in 5 seconds if it covered 14m during the 5th second find initial velocity and acceleration​

Answer 1

Body moving with uniform acceleration travels 50 m in 5 seconds.

Let a be the acceleration.

Let the initial velocity be u.

t be the time, t = 5 sec

s be the distance travelled in 5 sec , s = 50 m

According to second equation of motion,

$s = ut + \frac{1}{2} a {t}^{2} \\ \\ 50 = (u \times 5) + (\frac{1}{2} \times \: a \times {(5)}^{2} ) \\ \\ 50 = 5u + \frac{25}{2} a \\ \\ dividing \: by \: 5 \: on \: both \: sides \\ \\ 10 = u + \frac{5}{2} a \\ \\ 10 = \dfrac{2u + 5a}{2} \\ \\ 10 \times 2 = 2 u+ 5a \\ \\ 2 u+ 5a = 20$

Let the above equation be equation 1.

Body covers 14 m during the 5th second.

Distance covered during nth second, d = 14 m

n = 5 ; n is the duration

Distance covered in nth second,

$d = u + \dfrac{a}{2} (2n - 1) \\ \\ 14 = u + \dfrac{a}{2} ((2 \times 5) - 1) \\ \\ 14 = u + \dfrac{a}{2} (10 - 1) \\ \\ 14 = u + \frac{a}{2} (9) \\ \\ 14 = u + \frac{9a}{2} \\ \\ 14 = \dfrac{2u + 9a}{2} \\ \\ 14 \times 2 = 2 u+ 9a \\ \\ 2u + 9a = 28$

Let the above equation be equation 2.

Subtracting equation 1 from equation 2.

$2u + 9a - (2u + 5a) = 28 - 20 \\ \\ 2u + 9a - 2u - 5a = 8 \\ \\ 9a - 5a = 8 \\ \\ 4a = 8 \\ \\ a = \dfrac{8}{4} \\ \\ a = 2 \: \dfrac{m}{ {s}^{2} }$

The acceleration is 2 m/s².

Substitute the value of a in any of the equation.

Let's put a = 2 in equation 2.

$2u + 9a = 28 \\ \\ 2u + (9\times 2)= 28 \\ \\ 2u + 18 = 28 \\ \\ 2u = 28 -18 \\ \\ 2u = 10 \\ \\ u = \frac{10}{2} \\ \\ u = 5 \: \frac{m}{s}$

The initial velocity is 5 m/s.