Answer:
The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.
Explanation:
Given that,
Initial speed = 10.0 m/s
Acceleration = 3.00 m/s^2
Distance [tex]d = \dfrac{1}{4}\times1.609[/tex]
[tex]d = 0.40225\ km=0.40225\times10^{3}\ m[/tex]
We need to calculate the speed of the car,
Using equation of motion
[tex]v^2-u^2=2as[/tex]
Where, u = initial velocity
v = final velocity
a = acceleration
Put the value in the equation
[tex]v^2=(10.0)^2+2\times3.00\times0.40225\times10^{3}[/tex]
[tex]v^2=2513.5[/tex]
[tex]v=50.13\ m/s[/tex]
(B). We need to calculate the stopping distance of the car,
Using equation of motion again
[tex]v^2=u^2+2as[/tex]
Here,initial velocity = 50.13 m/s
Final velocity = 0
Acceleration = -4.50 m/s²
Put the value in the equation
[tex]0=(50.13)^2-2\times4.50\times s[/tex]
[tex]s=\dfrac{(50.13)^2}{2\times4.50}[/tex]
[tex]s=279.22\ m[/tex]
Hence, The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.
