A = 1.15m/s2, Vf = 80.0km/h --> we need it in m/s, so:
Vf = 80km/h × 1000m/1km × 1h/3600s
= 22.22m/s
Top speed = Vf, initial speed = Vi
time (t) = V(Vf-Vi) ÷ a
t = (22.22-0)m/s ÷ 1.15m/s2
t = 22.22m/s × s2/1.15m
= 19.32 seconds
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Answer:
Vf= 3.435 m/s
Explanation:
Given:
Initial velocity Vi =0 m/s (starting from Rest position)
θ = 37⁰
Distance S = 1 m
To find: Final Velocity Vf=?
fist we have to find the down slope net acceleration a = g sin θ
a= 9.81 sin 37⁰ = 5.9 m/s²
By 3rd equation of motion
2 a S= Vf² - Vi²
Vf = Square root ( 2 × 5.9 m/s² × 1 + 0 m/s)
Vf = Square root (11.8)
Vf= 3.435 m/s
Given parameters:
Speed of the car = 50m/s
Time = 10s
Unknown:
Distance traveled before the car stopped = ?
We need to establish speed - time - distance relationship.
Speed is the rate of change of distance with time. It is a scalar quantity .
Mathematically;
Speed =
Distance = speed x time
Input the parameters and solve;
Distance = 50 x 10 = 500m
The car traveled 500m before it stopped.