NT
Okay, so I need to figure out how to solve this problem where a car is moving at a constant speed and then the driver applies the brakes, causing the car to decelerate uniformly until it stops. The question is asking for the distance the car travels during the braking phase. Let me start by recalling what I know about motion under constant acceleration, which in this case is deceleration.
First, the car is moving at a constant speed initially. Let me denote that initial speed as ( v_0 ). When the driver applies the brakes, the car decelerates uniformly. Uniform deceleration means the acceleration is constant, right? So the acceleration ( a ) here would be negative because it's deceleration. The car comes to a stop, so the final velocity ( v ) would be 0 m/s.
I remember there are some kinematic equations that relate initial velocity, final velocity, acceleration, and displacement. The one that might be useful here is the equation that doesn't involve time, since the problem doesn't mention time. Let me try to recall that equation. I think it's ( v^2 = v_0^2 + 2a s ), where ( s ) is the displacement or distance traveled. Let me check if that's correct. Yes, that seems right. So if I rearrange this equation to solve for ( s ), I get ( s = \frac{v^2 - v_0^2}{2a} ). But since the car is decelerating, ( a ) is negative, and the final velocity ( v ) is 0. Plugging those values in, the equation becomes ( s = \frac{0 - v_0^2}{2a} ), which simplifies to ( s = -\frac{v_0^2}{2a} ). But since ( a ) is negative (deceleration), the negatives will cancel out, giving a positive distance, which makes sense.
Wait, let me make sure I'm not mixing up the signs here. If the car is decelerating, the acceleration is opposite to the direction of motion. So if we take the initial direction of motion as positive, then acceleration ( a ) is negative. So substituting ( v = 0 ), ( v_0 ) is positive, and ( a ) is negative into the equation ( v^2 = v_0^2 + 2a s ), we get ( 0 = v_0^2 + 2a s ). Solving for ( s ), we have ( s = -\frac{v_0^2}{2a} ). Since ( a ) is negative, the denominator becomes positive, so the whole expression is positive, which is correct because distance can't be negative.
Alternatively, maybe I can use another kinematic equation. Let's see. Another one is ( s = vt + \frac{1}{2} a t^2 ), but that requires knowing the time ( t ) it takes to stop, which we don't have. However, we can find ( t ) using ( v = v_0 + a t ). Since ( v = 0 ), solving for ( t ) gives ( t = -\frac{v_0}{a} ). Then substituting this into the displacement equation: ( s = v_0 (-\frac{v_0}{a}) + \frac{1}{2} a (-\frac{v_0}{a})^2 ). Let's compute that. The first term is ( -\frac{v_0^2}{a} ), and the second term is ( \frac{1}{2} a \times \frac{v_0^2}{a^2} = \frac{1}{2} \frac{v_0^2}{a} ). So combining these terms: ( -\frac{v_0^2}{a} + \frac{1}{2} \frac{v_0^2}{a} = -\frac{1}{2} \frac{v_0^2}{a} ). But since ( a ) is negative, this becomes positive, which matches the previous result. So both methods give the same answer, which is reassuring.
Therefore, the distance traveled during braking is ( \frac{v_0^2}{2 |a|} ), where ( |a| ) is the magnitude of the deceleration. But in terms of the given variables, if we stick to the original equation ( s = -\frac{v_0^2}{2a} ), since ( a ) is negative, this becomes positive. So the key formula here is ( s = \frac{v_0^2}{2 |a|} ).
Let me check the units to make sure. If ( v_0 ) is in meters per second (m/s) and ( a ) is in meters per second squared (m/s²), then the units for ( s ) would be (m²/s²) / (m/s²) = meters, which is correct. So the units check out.
Is there another way to think about this problem? Maybe using energy? If the car is braking, the work done by the brakes should equal the kinetic energy lost by the car. The work done is force times distance. The force here is the braking force, which is mass times acceleration (Newton's second law, ( F = ma )). So work done ( W = F \times s = ma \times s ). The kinetic energy lost is ( \frac{1}{2} m v_0^2 ) since the final kinetic energy is zero. Setting work done equal to kinetic energy lost: ( ma s = \frac{1}{2} m v_0^2 ). The mass ( m ) cancels out, giving ( a s = \frac{1}{2} v_0^2 ), so ( s = \frac{v_0^2}{2a} ). Wait, but here ( a ) is the deceleration, which is a positive value if we consider the magnitude. However, in the kinematic equations earlier, ( a ) was negative. So this energy approach gives the same formula but with ( a ) as a positive magnitude. So in this case, ( s = \frac{v_0^2}{2a} ), where ( a ) is the magnitude of the deceleration. That's consistent with the earlier result when considering absolute values.
Therefore, regardless of the method—kinematic equations or energy conservation—we arrive at the same formula for the stopping distance. This cross-verification makes me more confident that the formula is correct.
Now, let me consider if there are any common mistakes or pitfalls here. One might be mixing up the sign of acceleration. If someone forgets that deceleration is negative in the kinematic equations and uses a positive ( a ), they would end up with a negative distance, which is impossible. So it's crucial to keep track of the direction of acceleration relative to the initial velocity.
Another possible mistake is confusing speed and velocity. Since the problem mentions constant speed and uniform deceleration, we can treat this as one-dimensional motion, so speed and velocity have the same magnitude here. But in cases where motion is two-dimensional, that might not hold. However, in this problem, it's straightforward.
Also, ensuring that all units are consistent is important. For example, if the initial speed is given in kilometers per hour and acceleration in meters per second squared, we need to convert them to the same unit system before plugging into the formula.
Let me test the formula with some example values to see if it makes sense. Suppose a car is moving at 10 m/s and decelerates at 2 m/s². Plugging into the formula: ( s = \frac{10^2}{2 \times 2} = \frac{100}{4} = 25 ) meters. That seems reasonable. If the car is going faster, say 20 m/s (72 km/h) with the same deceleration, the stopping distance would be ( \frac{400}{4} = 100 ) meters, which also makes sense—higher speed means longer stopping distance with the same deceleration.
Alternatively, if the deceleration is higher, say 5 m/s², then stopping distance from 10 m/s would be ( \frac{100}{10} = 10 ) meters. So more aggressive braking reduces stopping distance, which is intuitive.
Therefore, the formula behaves as expected in different scenarios, which adds to its validity.
In summary, the key steps are:
Identify initial velocity ( v_0 ), final velocity ( v = 0 ), and constant deceleration ( a ).
Use the kinematic equation ( v^2 = v_0^2 + 2a s ).
Solve for ( s ), remembering that ( a ) is negative, leading to ( s = \frac{v_0^2}{2 |a|} ).
Verify using energy considerations or alternative kinematic equations.
Check units and test with example values to ensure the formula is applied correctly.
I think that covers all aspects. The main takeaway is that stopping distance depends on the square of the initial speed and inversely on the deceleration. This is why higher speeds result in significantly longer stopping distances, which is a critical safety consideration in driving.
To determine the distance a car travels during braking when decelerating uniformly from an initial speed ( v_0 ) to a stop, we use the kinematic equation:
[
v^2 = v_0^2 + 2a s
]
Here:
( v = 0 ) (final velocity, since the car stops),
( v_0 ) = initial velocity,
( a ) = constant deceleration (negative value),
( s ) = stopping distance.
Step-by-Step Solution:
https://www.18122.com/url/usdt/usdt1836.jpg
Rearrange the equation to solve for ( s ):
[
0 = v_0^2 + 2a s \implies s = -\frac{v_0^2}{2a}
]
Interpret the sign of ( a ):
Since ( a ) is deceleration (opposite to motion), substitute ( |a| ) (magnitude of ( a )):
[
s = \frac{v_0^2}{2 |a|}
]
Energy Conservation Verification:
Kinetic energy lost equals work done by braking force:
[
\frac{1}{2} m v_0^2 = F \cdot s = (m |a|) \cdot s \implies s = \frac{v_0^2}{2 |a|}
]
Final Formula:
[
\boxed{s = \frac{v_0^2}{2 |a|}}
]
Key Points:
Stopping distance is proportional to the square of initial speed.
It is inversely proportional to the magnitude of deceleration.
Ensure consistent units (e.g., ( v_0 ) in m/s, ( |a| ) in m/s²).
Example:
If ( v_0 = 10 , \text{m/s} ) and ( |a| = 2 , \text{m/s}^2 ):
[
s = \frac{10^2}{2 \times 2} = 25 , \text{meters}.
]
This method confirms that safe braking distances increase significantly with speed and decrease with stronger deceleration.
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