Answer:
1.5 m/s
32.25 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
a = Acceleration
s = Displacement
Equation of motion
[tex]v=u+at\\\Rightarrow v=0+1.5\times 5\\\Rightarrow v=7.5\ m/s[/tex]
The velocity of the car at the end of the acceleration period is 7.5 m/s
This velocity will be considered as the initial velocity of the decelerating stage
[tex]v=u+at\\\Rightarrow v=7.5-2\times 3\\\Rightarrow v=1.5\ m/s[/tex]
The velocity of the car at the end of the deceleration period is 1.5 m/s
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 1.5\times 5^2\\\Rightarrow s=18.75\ m[/tex]
Distance traveled in the acceleration period is 18.75 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=7.5\times 3+\frac{1}{2}\times -2\times 3^2\\\Rightarrow s=13.5\ m[/tex]
Distance traveled in the deceleration period is 13.5 m
Total distance traveled is 18.75+13.5 = 32.25 m
