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Sagot :
To solve the problem of finding the acceleration when the car stops after traveling a certain distance, we use the kinematic equations that describe motion under constant acceleration.
Given:
- Initial velocity, [tex]\( u = 9 \, \text{m/s} \)[/tex]
- Stopping distance, [tex]\( s = 9 \, \text{m} \)[/tex]
- Final velocity, [tex]\( v = 0 \, \text{m/s} \)[/tex] (since the car stops)
We can use the following kinematic equation to find the acceleration:
[tex]\[ v^2 = u^2 + 2as \][/tex]
First, let's plug in the known values:
[tex]\[ 0^2 = 9^2 + 2 \cdot a \cdot 9 \][/tex]
Simplify the equation:
[tex]\[ 0 = 81 + 18a \][/tex]
To solve for [tex]\( a \)[/tex], isolate [tex]\( a \)[/tex] on one side of the equation:
[tex]\[ 18a = -81 \][/tex]
Now, divide both sides by 18 to find [tex]\( a \)[/tex]:
[tex]\[ a = \frac{-81}{18} \][/tex]
[tex]\[ a = -4.5 \, \text{m/s}^2 \][/tex]
Therefore, the acceleration of the car is [tex]\(-4.5 \, \text{m/s}^2\)[/tex]. The negative sign indicates that the car is decelerating.
Given:
- Initial velocity, [tex]\( u = 9 \, \text{m/s} \)[/tex]
- Stopping distance, [tex]\( s = 9 \, \text{m} \)[/tex]
- Final velocity, [tex]\( v = 0 \, \text{m/s} \)[/tex] (since the car stops)
We can use the following kinematic equation to find the acceleration:
[tex]\[ v^2 = u^2 + 2as \][/tex]
First, let's plug in the known values:
[tex]\[ 0^2 = 9^2 + 2 \cdot a \cdot 9 \][/tex]
Simplify the equation:
[tex]\[ 0 = 81 + 18a \][/tex]
To solve for [tex]\( a \)[/tex], isolate [tex]\( a \)[/tex] on one side of the equation:
[tex]\[ 18a = -81 \][/tex]
Now, divide both sides by 18 to find [tex]\( a \)[/tex]:
[tex]\[ a = \frac{-81}{18} \][/tex]
[tex]\[ a = -4.5 \, \text{m/s}^2 \][/tex]
Therefore, the acceleration of the car is [tex]\(-4.5 \, \text{m/s}^2\)[/tex]. The negative sign indicates that the car is decelerating.
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