Answer :
(a) The acceleration is, \(0.278m/s^2\)
(b) Distance it travels during this time is, 156.225 m
Solution for part (a) :
Formula used :
\(a=\frac{v-u}{t}\)
where,
a = acceleration
v = final velocity = \(45Km/hr=45\times \frac{5}{18}=12.5m/s\)
u = initial velocity = \(30Km/hr=30\times \frac{5}{18}=8.33m/s\)
t = time = 15 s
Now put all the given values in the above formula, we get
\(a=\frac{(12.5-8.33)m/s}{15s}=0.278m/s^2\)
Therefore, the acceleration is, \(0.278m/s^2\)
Solution for part (b) :
Formula used :
\(s=u\times t+\frac{1}{2}a\times t^2\)
where,
s = distance traveled
Now put all the given values in this formula, we get
\(s=(8.33m/s)\times (15s)+\frac{1}{2}\times (0.278m/s^2)\times (15s)^2=156.225m\)
Therefore, the distance it travels during this time is, 156.225 m
