Answer:
(a) 21 m/s
(b) 15.78 m/s
(c) 12.5 m/s
Explanation:
Given;
Initial velocity, u = +7.0 m/s
accelerates, a = +0.80 m/s²
distance, s = 245 m
One of equation of motions is given by;
[tex]v^{2} = u^{2} + 2as[/tex]
where,
v is the final velocity in m/s
u is the initial velocity in m/s
a is the acceleration in m/s²
s is the distance in m
(a) The velocity of the car at end of the acceleration will be when it covers 245 m
[tex]v^{2} = 7^{2} + (2 \ * 0.80s \ * 245)[/tex]
[tex]v^{2} = 49 + 392[/tex]
[tex]v^{2} = 441[/tex]
[tex]v = \sqrt{441}[/tex]
[tex]v = 21 \ m/s[/tex]
(b) Velocity after the car accelerates for 125 m
[tex]v^{2} = 7^{2} + (2 \ * 0.80s \ * 125)[/tex]
[tex]v^{2} = 49 + 200[/tex]
[tex]v^{2} = 249[/tex]
[tex]v = \sqrt{249}[/tex]
[tex]v = 15.78 \ m/s[/tex]
(c) Velocity after the car accelerates for 67 m
[tex]v^{2} = 7^{2} + (2 \ * 0.80s \ * 67)[/tex]
[tex]v^{2} = 49 + 107.2[/tex]
[tex]v^{2} = 156.2[/tex]
[tex]v = \sqrt{156.2}[/tex]
[tex]v = 12.5 \ m/s[/tex]
