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 the toy car travels while accelerating, we use the kinematic equation:
[
v^2 = u^2 + 2as
]
Where:
- ( v = 1.8 ) m/s (final velocity),
- ( u = 0 ) m/s (initial velocity, since it starts from rest),
- ( a = 0.40 ) m/s² (acceleration),
- ( s ) is the distance traveled.
Step 1: Solve for ( s )
[
(1.8)^2 = (0)^2 + 2(0.40)s
]
[
3.24 = 0.80s
]
[
s = \frac{3.24}{0.80}
]
[
s = 4.05 \text{ m}
]
Final Answer:
The toy car travels 4.05 meters before reaching 1.8 m/s.
Explanation:
The problem involves a toy car accelerating uniformly from rest. Since it starts from zero velocity and increases speed at a constant rate (0.40 m/s²), we can use kinematic equations to determine how far it moves before reaching the target velocity (1.8 m/s).
First, we recognize that we are not given the time, but we do have acceleration and final velocity. This suggests using the kinematic equation:
[
v^2 = u^2 + 2as
]
This equation relates velocity, acceleration, and displacement without needing time. Since the toy car starts from rest (( u = 0 )), the equation simplifies, making it easy to solve for ( s ).
After plugging in the values, we find that the toy car moves 4.05 meters before reaching the given speed. This makes sense because with a small acceleration of 0.40 m/s², it takes some distance to gradually reach a higher velocity.
This problem demonstrates the importance of kinematics in predicting motion, showing how we can determine displacement without knowing time, as long as acceleration and velocity are given.