To determine the final velocity of the toy car under the influence of the unbalanced force, we can use Newton's second law of motion and the kinematic equations.
First, we calculate the acceleration produced by the unbalanced force using F = ma, where \( F \) is the force, \( m \) is the mass of the car, and \( a \) is the acceleration.
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Given data:
- Mass of the car, \( m = 1.0 , \text{kg} \)
- Initial velocity, \( u = 5.0 , \text{m/s} \)
- Force, \( F = 2.0 , \text{N} \)
- Time interval, \( t = 4.0 , \text{s} \)
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Calculate acceleration: \[ a = \frac{F}{m} = \frac{2.0 , \text{N}}{1.0 , \text{kg}} = 2.0 , \text{m/s}^2 \]
Now we can find the final velocity \( v \) after the time interval using the equation: \[ v = u + at \]
(a) If the force acts in the direction of motion of the car:
When the force acts in the same direction as the motion:
- Acceleration \( a = 2.0 , \text{m/s}^2 \)
\[ v = 5.0 , \text{m/s} + (2.0 , \text{m/s}^2 \times 4.0 , \text{s}) \] \[ v = 5.0 , \text{m/s} + 8.0 , \text{m/s} = 13.0 , \text{m/s} \]
Final velocity (a):
\[ v = 13.0 , \text{m/s} \]
(b) If the force acts in the opposite direction to the motion of the car:
When the force acts opposite to the direction of motion:
- The acceleration will be negative, \( a = -2.0 , \text{m/s}^2 \)
\[ v = 5.0 , \text{m/s} + (-2.0 , \text{m/s}^2 \times 4.0 , \text{s}) \] \[ v = 5.0 , \text{m/s} - 8.0 , \text{m/s} = -3.0 , \text{m/s} \]
Final velocity (b):
\[ v = -3.0 , \text{m/s} \]
This indicates that the car has reversed its direction of motion after being acted upon by the force in the opposite direction.
Summary:
- (a) Final velocity with force in the direction of motion: 13.0 m/s
- (b) Final velocity with force opposite to the motion: -3.0 m/s