Answer:
[tex]\boxed {\boxed {\sf 2.2 \ m/s^2}}[/tex]
Explanation:
We are asked to solve for the magnitude of the car's acceleration.
We are given the initial speed, final speed, and distance, so we will use the following kinematic equation.
[tex]{v_f}^2={v_i}^2+2ad[/tex]
The car is initially traveling at 15.0 meters per second and accelerates to 20.0 meters per second over a distance of 40.0 meters. Therefore,
Substitute the values into the formula.
[tex](20.0 \ m/s)^2= (15.0 \ m/s)^2 + 2 a (40.0 \ m)[/tex]
Solve the exponents.
[tex]400.0 \ m^2/s^2 = 225.0 \ m^2/s^2 + 2 a(40.0 \ m)[/tex]
Subtract 225.0 m²/s² from both sides of the equation.
[tex]400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 225.0 \ m^2/s^2 -225 \ m^2/s^2 +2a(40.0 \ m)[/tex]
[tex]400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 2a(40.0 \ m)[/tex]
[tex]175 \ m^2/s^2 = 2a(40.0 \ m)[/tex]
Multiply on the right side of the equation.
[tex]175 \ m^2/s^2 =80.0 \ m *a[/tex]
Divide both sides by 80.0 meters to isolate the variable a.
[tex]\frac {175 \ m^2/s^2}{80.0 \ m}= \frac{80.0 \ m *a}{80.0 \ m}[/tex]
[tex]\frac {175 \ m^2/s^2}{80.0 \ m}=a[/tex]
[tex]2.1875 \ m/s^2 =a[/tex]
Round to the tenths place. The 8 in the hundredth place tells us to round the 1 up to a 2.
[tex]2.2 \ m/s^2=a[/tex]
The magnitude of the car's acceleration is 2.2 meters per second squared.
