Question

the distance moved by a body at the end of t1=3s and t2=5s are 126m and 250m respectively. what is the initial velocity of the body.

Answer 1

$s(t) = v_0t +\frac{at^2}{2}$

$s(3)= 3v_0+4.5a =126(1)$

$s(5)= 5v_0+12.5a = 250(2)$

$\text{From (1) and (2)}$

$v_0=27\frac{m}{s}$

$\text{Answer:} 27\frac{m}{s}$