a woman driving at a speed of 23 m/s sees a deer on the road ahead and applies the brakes when she is 210m from the deer. if the deer does not move and the car stops right before it hits the deer, what is the acceleration provided by the car's brakes?
V^2 = Vo^2 + 2a*d.
a = (V^2-Vo^2)/2d
a = (0-(23)^2)/420 = 1.26 m/s^2.
Well, first off, I'm glad the woman didn't turn into a "deer" in headlights! Now, let's get calculating. We have the initial velocity of the car (23 m/s), the displacement (210 m), and we want to find the acceleration. To solve this, we can use one of the equations of motion. Let's go with the one that relates velocity, acceleration, and displacement: v^2 = u^2 + 2as. Since the car stops completely, its final velocity is 0 m/s. Plugging the given values, we have: 0 = (23 m/s)^2 + 2a(210 m). Solving for acceleration, we get: a = - (23 m/s)^2 / (2 * 210 m). Now, let's not be negative about it, as it represents the deceleration. So, the acceleration provided by the car's brakes is approximately -0.767 m/s^2. And just like that, it's a braking bad situation for the car!
To find the acceleration provided by the car's brakes, we can use the kinematic equation:
\(v_f^2 = v_i^2 + 2a \cdot d\)
Where:
\(v_f\) = final velocity (0 m/s because the car stops)
\(v_i\) = initial velocity (23 m/s)
\(a\) = acceleration (what we need to find)
\(d\) = distance (210 m)
Now let's plug in the given values and solve for acceleration:
\(0^2 = 23^2 + 2a \cdot 210\)
Simplifying the equation:
\(0 = 529 + 420a\)
Rearranging the equation:
\(420a = -529\)
Finally, solving for \(a\):
\(a = \frac{-529}{420}\)
\(a \approx -1.26\) m/s²
Therefore, the acceleration provided by the car's brakes is approximately -1.26 m/s². The negative sign indicates that the acceleration is in the opposite direction to the initial motion of the car.
To find the acceleration provided by the car's brakes, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, since the car stops)
u = initial velocity (23 m/s)
a = acceleration (unknown)
s = distance (210 m)
First, let's rearrange the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Substituting the given values into the equation:
a = (0^2 - 23^2) / (2 × 210)
Simplifying the equation:
a = (-529) / 420
a ≈ -1.26 m/s^2
So, the acceleration provided by the car's brakes is approximately -1.26 m/s^2. The negative sign indicates deceleration, which is expected since the car is slowing down.