Answer :
To solve the problem of how long it takes for a 1330 kg car, initially at rest, to move a distance of 5.00 meters when a force of 187 N is applied, we will follow a step-by-step process. We will use Newton's second law and kinematic equations to find the time.
Step 1: Calculate the acceleration.
According to Newton's second law, the force applied on an object is equal to the mass of the object multiplied by its acceleration [tex]\( (F = ma) \)[/tex].
Given:
- Mass of the car ([tex]\( m \)[/tex]) = 1330 kg
- Force applied ([tex]\( F \)[/tex]) = 187 N
We need to find the acceleration [tex]\( a \)[/tex]:
[tex]\[ a = \frac{F}{m} \][/tex]
Substitute the given values:
[tex]\[ a = \frac{187 \, \text{N}}{1330 \, \text{kg}} \][/tex]
[tex]\[ a \approx 0.1406 \, \text{m/s}^2 \][/tex]
Step 2: Use the kinematic equation.
We need to find the time taken to move the car 5.00 meters. We will use the following kinematic equation, which relates distance [tex]\( d \)[/tex], initial velocity [tex]\( v_0 \)[/tex], acceleration [tex]\( a \)[/tex], and time [tex]\( t \)[/tex]:
[tex]\[ d = v_0 t + \frac{1}{2} a t^2 \][/tex]
Given:
- Distance ([tex]\( d \)[/tex]) = 5.00 m
- Initial velocity ([tex]\( v_0 \)[/tex]) = 0 (since the car is initially at rest)
- Acceleration ([tex]\( a \)[/tex]) = 0.1406 m/s[tex]\(^2\)[/tex]
Since the initial velocity [tex]\( v_0 \)[/tex] is 0, the equation simplifies to:
[tex]\[ d = \frac{1}{2} a t^2 \][/tex]
Rearrange the equation to solve for time [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{2d}{a} \][/tex]
[tex]\[ t = \sqrt{\frac{2d}{a}} \][/tex]
Substitute the given values:
[tex]\[ t = \sqrt{\frac{2 \times 5.00 \, \text{m}}{0.1406 \, \text{m/s}^2}} \][/tex]
[tex]\[ t \approx 8.433 \, \text{s} \][/tex]
Therefore, it takes approximately 8.433 seconds for the car to move a distance of 5.00 meters when pushed with a force of 187 N.
Step 1: Calculate the acceleration.
According to Newton's second law, the force applied on an object is equal to the mass of the object multiplied by its acceleration [tex]\( (F = ma) \)[/tex].
Given:
- Mass of the car ([tex]\( m \)[/tex]) = 1330 kg
- Force applied ([tex]\( F \)[/tex]) = 187 N
We need to find the acceleration [tex]\( a \)[/tex]:
[tex]\[ a = \frac{F}{m} \][/tex]
Substitute the given values:
[tex]\[ a = \frac{187 \, \text{N}}{1330 \, \text{kg}} \][/tex]
[tex]\[ a \approx 0.1406 \, \text{m/s}^2 \][/tex]
Step 2: Use the kinematic equation.
We need to find the time taken to move the car 5.00 meters. We will use the following kinematic equation, which relates distance [tex]\( d \)[/tex], initial velocity [tex]\( v_0 \)[/tex], acceleration [tex]\( a \)[/tex], and time [tex]\( t \)[/tex]:
[tex]\[ d = v_0 t + \frac{1}{2} a t^2 \][/tex]
Given:
- Distance ([tex]\( d \)[/tex]) = 5.00 m
- Initial velocity ([tex]\( v_0 \)[/tex]) = 0 (since the car is initially at rest)
- Acceleration ([tex]\( a \)[/tex]) = 0.1406 m/s[tex]\(^2\)[/tex]
Since the initial velocity [tex]\( v_0 \)[/tex] is 0, the equation simplifies to:
[tex]\[ d = \frac{1}{2} a t^2 \][/tex]
Rearrange the equation to solve for time [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{2d}{a} \][/tex]
[tex]\[ t = \sqrt{\frac{2d}{a}} \][/tex]
Substitute the given values:
[tex]\[ t = \sqrt{\frac{2 \times 5.00 \, \text{m}}{0.1406 \, \text{m/s}^2}} \][/tex]
[tex]\[ t \approx 8.433 \, \text{s} \][/tex]
Therefore, it takes approximately 8.433 seconds for the car to move a distance of 5.00 meters when pushed with a force of 187 N.
