(a) Suppose a body has an initial velocity ‘u’ and a uniform acceleration’ a’ for time ‘t’ so that its final velocity becomes ‘v’.

Let the distance travelled by the body in this time be ‘s’.

The distance travelled by a moving body in time ‘t’ can be found out by considering its average velocity. Since the initial velocity of the body is ‘u’ and its final velocity is ‘v’, the average velocity is given by:

Average velocity = \(\frac{Initial\;velocity + Final\;velocity}{2}\)

That is, Average velocity = \(\frac{u + v}{2}\)

Also, Distance travelled = Average velocity x Time

So, s = \((\frac{u + v}{2}) \times t\)   ......(1)

From the first equation of motion, we have, 

v = u + at

Put this value of v in equation (1), we get,

s = \((\frac{u + u + at}{2}) \times t\)

or s = \(\frac{(2u + at) \times t}{2}\)

or s = \(\frac{2ut + at^2}{2}\)

or s = ut + \(\frac{1}{2}at^2\)

where, s = distance travelled by the body in time t

u = Initial velocity of the body

and a = acceleration

Acceleration = \(\frac{Final\;velocity - Initial\;velocity}{Time\;taken}\)

So, a = \(\frac{v - u}{t}\)

= \(\frac{10 - 0}{600}\) = \(\frac{10}{600}\) m/s²

= \(\frac{1}{60}\) m/s²

= 0.016 m/s²

(b) Initial velocity, u = 0 m/s

Final velocity, v = 36km/h = 10 m/s

Time, t = 10 min = 10 x 60 = 600 sec