(a) A boy runs in a line and his displacement at time t seconds after leaving the start point O is X metres, where 20X = 4t\(^2\) + t\(^3\). Find the:
(i) velocity of the body when t = 15 seconds (ii) value of t for which the acceleration of the body is 8 times his initial acceleration
(b) A body of mass 6 kg moves with a velocity of 7 ms\(^{-1}\). It collides with a second body moving in the opposite direction with a velocity of 5 ms\(^{-1}\). After collision, the two bodies move together with a velocity of 4 ms\(^{-1}\). Find the mass of the second body.
Explanation
(a)i X = \(\frac{1}{5}\)t\(^2\) + \(\frac{1}{20}\)t\(^3\) ( after dividing thru by 20)
\(\frac{dX}{dt}\) = velocity = \(\frac{2}{5}\)t + \(\frac{3}{20}\))t\(^2\)
at t = 15, v = \(\frac{2}{5}\)(15) + \(\frac{3}{20}\)(15)\(^2\) = 6 + 33.75 = 39.75m/s
Acceleration = \(\frac{dV}{dt}\) = \(\frac{2}{5}\)t + \(\frac{3}{20}\)t\(^2\) = \(\frac{2}{5}\) + \(\frac{3}{10}\)t
for initial acceleration, t = 0
a = \(\frac{2}{5}\)ms\(^2\)
(8) \(\frac{2}{5}\) = \(\frac{2}{5}\) + \(\frac{3}{10}\)t
40 + 30t = 320
30t = 320 - 40 = 280
t = \(\frac{280}{30}\) = 9\(\frac{1}{3}\)secs.
(b) m\(_1\)u\(_1\) + m\(_2\)u\(_2\) = v\(_c\)(m\(_1\) + m\(_2\))
m\(_1\) = 6kg, u\(_1\) = 7m/s, m\(_2\) = ?, u\(_2\) = - 5 m/s
6(7) + m\(_2\)(-5) = 4(6 + m\(_2\))
42 - 5m\(_2\) = 24 + 4m\(_2\)
42 - 24 = 4m\(_2\) + 5m\(_2\)
9m\(_2\) = 18
m\(_2\) = \(\frac{18}{2}\) = 2 kg.