Question

A car initially traveling at a speed of 15.0 m/s accelerates uniformly to a speed of 20.0 m/s over a distance of 40.0 meters. What is the magnitude of the car's acceleration?

Answers:
1.1 m/s^2
2.0 m/s^2
2.2 m/s^2
9 m/s^2

Answer 1

`\boxed {\boxed {\sf 2.2 \ m/s^2}}`

Step-by-step 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.

`{v_f}^2={v_i}^2+2ad`

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,

• 
  `v_f`= 20.0 m/s
• 
  `v_i`= 15.0 m/s
• d= 40.0 m

Substitute the values into the formula.

`(20.0 \ m/s)^2= (15.0 \ m/s)^2 + 2 a (40.0 \ m)`

Solve the exponents.

• (20.0 m/s)² = 20.0 m/s * 20.0 m/s = 400.0 m²/s²
• (15.0 m/s)² = 15.0 m/s * 15.0 m/s = 225.0 m²/s²

`400.0 \ m^2/s^2 = 225.0 \ m^2/s^2 + 2 a(40.0 \ m)`

Subtract 225.0 m²/s² from both sides of the equation.

`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)`

`400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 2a(40.0 \ m)`

`175 \ m^2/s^2 = 2a(40.0 \ m)`

Multiply on the right side of the equation.

`175 \ m^2/s^2 =80.0 \ m *a`

Divide both sides by 80.0 meters to isolate the variable a.

`\frac {175 \ m^2/s^2}{80.0 \ m}= (80.0 \ m *a)/(80.0 \ m)`

`\frac {175 \ m^2/s^2}{80.0 \ m}=a`

`2.1875 \ m/s^2 =a`

Round to the tenths place. The 8 in the hundredth place tells us to round the 1 up to a 2.

`2.2 \ m/s^2=a`

The magnitude of the car's acceleration is 2.2 meters per second squared.