Answer: see below
Explanation:
The intial speed is u=0[tex]\frac{m}{s}[/tex]
The acceleration is a= 3.3[tex]\frac{m}{s}^2[/tex]
The distance is s=60m
Apply the equation of motion
[tex]v^{2}[/tex]=[tex]u^{2}[/tex]+2as
So,
[tex]v^{2}[/tex]=0+2*3.3*60=396
v=[tex]\sqrt{396}[/tex] = 19.89 m/s ≈20 m/s
Applying the equation of motion
v=u+at
t= [tex]\frac{v-u}{a}[/tex] =[tex]\frac{19.89-0}{3.3}[/tex] = 6.02seconds
