Determine the stopping distances for a car with an initial speed of 87 km/h and human reaction time of 2.0 s for the following accelerations.
(a) a = -4.0 m/s2
(b) a = -8.0 m/s2
Vo = 87km/h = 87,000m/3600s = 24.2 m/s. = Initial velocity.
a. d1 = 24.2 * 2 =
V^2 = Vo^2 + 2a*d2.
V = 0, Vo = 24.2 m/s, a = -4.0 m/s^2, d2 = ?.
d1+d2 = Stopping distance.
b. V^2 = Vo^2 + 2a*d2.
V = 0, Vo = 24.2 m/s^2, a = -8.0 m/s^2, d2 = ?.
d1+d2 = stopping distance.
To determine the stopping distances for a car with an initial speed of 87 km/h and a reaction time of 2.0 s for different accelerations, we will use the following formula:
Stopping Distance = Initial Velocity × Reaction Time + 0.5 × Acceleration × Reaction Time^2
First, let's convert the initial velocity from km/h to m/s:
Initial Velocity = 87 km/h = 87 × (1000 m / 3600 s) = 24.17 m/s
(a) For an acceleration of -4.0 m/s^2:
Stopping Distance = 24.17 m/s × 2.0 s + 0.5 × (-4.0 m/s^2) × (2.0 s)^2
Stopping Distance = 48.34 m + (-8.0 m/s^2) × 4.0 s^2
Stopping Distance = 48.34 m - 16.0 m
Stopping Distance = 32.34 m
Therefore, the stopping distance for a car with an initial speed of 87 km/h, a reaction time of 2.0 s, and an acceleration of -4.0 m/s^2 is 32.34 m.
(b) For an acceleration of -8.0 m/s^2:
Stopping Distance = 24.17 m/s × 2.0 s + 0.5 × (-8.0 m/s^2) × (2.0 s)^2
Stopping Distance = 48.34 m + (-16.0 m/s^2) × 4.0 s^2
Stopping Distance = 48.34 m - 32.0 m
Stopping Distance = 16.34 m
Therefore, the stopping distance for a car with an initial speed of 87 km/h, a reaction time of 2.0 s, and an acceleration of -8.0 m/s^2 is 16.34 m.
To determine the stopping distances for a car with different accelerations, we can use the kinematic equation:
\[ d = v_i t + \frac{1}{2} a t^2 \]
where:
- \(d\) is the stopping distance,
- \(v_i\) is the initial velocity (convert km/h to m/s),
- \(a\) is the constant acceleration,
- \(t\) is the human reaction time.
Let's solve this equation for each case:
(a) Acceleration \(a = -4.0 \, \text{m/s}^2\):
- Convert the initial velocity from km/h to m/s:
\( v_i = 87 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} \)
\( v_i = 24.17 \, \text{m/s} \)
- Calculate the stopping distance:
\( d = (24.17 \, \text{m/s})(2.0 \, \text{s}) + \frac{1}{2} (-4.0 \, \text{m/s}^2)(2.0 \, \text{s})^2 \)
(b) Acceleration \(a = -8.0 \, \text{m/s}^2\):
- Convert the initial velocity from km/h to m/s:
\( v_i = 87 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} \)
\( v_i = 24.17 \, \text{m/s} \)
- Calculate the stopping distance:
\( d = (24.17 \, \text{m/s})(2.0 \, \text{s}) + \frac{1}{2} (-8.0 \, \text{m/s}^2)(2.0 \, \text{s})^2 \)
Now, plug in the values and calculate the distances for each case.