92. A car traveling at 20 m/s applies brakes, causing a deceleration of $5 \, \text{m/s}^2$. If the mass of the car is 1000 kg, what is the stopping distance considering both frictional force and air pressure effects?
Key Concept: Understanding Forces and Motion, Pressure and Its Effects
b) 40 meters
[Solution Description]
First, calculate the decelerating force using Newton's second law: $F = ma = 1000 \, \text{kg} \times 5 \, \text{m/s}^2 = 5000 \, \text{N}$.
Next, use the kinematic equation to find stopping distance: $v^2 = u^2 + 2as$, where final velocity $v = 0$, initial velocity $u = 20 \, \text{m/s}$, and acceleration $a = -5 \, \text{m/s}^2$. Rearranged: $0 = (20)^2 + 2(-5)s$ leads to $400 = 10s$, so $s = 40 \, \text{m}$.
The stopping distance is primarily determined by these factors, with minor adjustments for air pressure being negligible in this scenario.
Your Answer is correct.
b) 40 meters
[Solution Description]
First, calculate the decelerating force using Newton's second law: $F = ma = 1000 \, \text{kg} \times 5 \, \text{m/s}^2 = 5000 \, \text{N}$.
Next, use the kinematic equation to find stopping distance: $v^2 = u^2 + 2as$, where final velocity $v = 0$, initial velocity $u = 20 \, \text{m/s}$, and acceleration $a = -5 \, \text{m/s}^2$. Rearranged: $0 = (20)^2 + 2(-5)s$ leads to $400 = 10s$, so $s = 40 \, \text{m}$.
The stopping distance is primarily determined by these factors, with minor adjustments for air pressure being negligible in this scenario.