A toy car is released from rest. It accelerates at 0.40 m/s2 until it reaches a speed of 1.8 m/s. How far does it travel in this time?
The correct answer and explanation is :
To find the distance traveled by the toy car, we can use kinematic equations. Given:
- Initial velocity, ( u = 0 ) m/s (since it starts from rest)
- Final velocity, ( v = 1.8 ) m/s
- Acceleration, ( a = 0.40 ) m/s²
- Distance, ( s ) (to be found)
We use the kinematic equation:
[
v^2 = u^2 + 2as
]
Substituting the given values:
[
(1.8)^2 = (0)^2 + 2(0.40)s
]
[
3.24 = 0.8s
]
[
s = \frac{3.24}{0.8}
]
[
s = 4.05 \text{ m}
]
Explanation (300 words)
To solve this problem, we applied the kinematic equation that relates velocity, acceleration, and displacement. The equation:
[
v^2 = u^2 + 2as
]
is useful when time is not given but we know acceleration and velocities.
Since the toy car starts from rest (( u = 0 )), the equation simplifies to:
[
v^2 = 2as
]
Rearranging for ( s ):
[
s = \frac{v^2}{2a}
]
This formula gives us the total distance traveled during the acceleration phase.
By substituting the given values, we found that the car travels 4.05 meters before reaching its maximum speed of 1.8 m/s.
Physically, this means that while the toy car accelerates at a rate of 0.40 m/s², it covers this distance before reaching the given speed. The acceleration is relatively small, so the car takes some distance to build up speed.
This concept is commonly seen in physics problems involving motion with constant acceleration. Similar calculations apply to real-world scenarios like cars accelerating from a stoplight or objects rolling down an incline under gravity.
Thus, the car moves 4.05 meters before reaching the final velocity of 1.8 m/s.