A 1200 kg car is moving at 95 km/h when the driver notices a deer down the road. He immediately moves his foot toward the brake pedal, taking only 0.50 seconds before the car starts slowing down. The brakes cause a net force of 2400 N (backwards) on the car for 2.0 seconds. The deer then jumps out of the way and the driver lifts his foot off the brake pedal. How far does the car move in the 2.5 seconds starting from when the driver sees the deer?
Using newtons law.
95 km/hr *1000/3600 = 26.4 m/s
in .5 seconds moves 13.2 meters
then
a = F/m = -2400/1200 = - 2 m/s^2
0 = Vo - a t
d = 13.2 + 26.4 t - (1/2) (2) t^2
t = 2
so
d = 13.2 + 48.8 = 62 meters
d = 13.2 +
Well, well, well! We have ourselves a physics problem, don't we? Let's see if I can add a little humor to this equation!
Alright, let's break this down step by step, just like the car breaking down (not literally, I hope).
First, let's calculate the initial velocity of the car when the driver sees the deer. We know that it's moving at 95 km/h, but we need to convert that to meters per second because we're dealing with Newton's laws here. So, quick conversion: v = 95 km/h * (1000 m/3600 s) = 26.39 m/s.
Now, we can use Newton's second law, which states that force (F) equals mass (m) times acceleration (a), to find the acceleration while braking. F = ma, rearranging it, we get a = F/m. Plugging in the numbers, we get a = 2400 N / 1200 kg = 2 m/s^2.
Since we're dealing with a constant acceleration, we can use the kinematic equation: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. Plug in the numbers, we get s = (26.39 m/s) * 2.0 s + (1/2)*(2 m/s^2)*(2.0 s)^2 = ...*calculating, calculating*... 56.78 meters!
Alright, we're almost there! Now, the driver lifts his foot off the brake pedal, and the car no longer has any force acting on it. Therefore, the car will simply continue at a constant velocity until the 2.5 seconds are up. The distance traveled during this time is simply u * t = (26.39 m/s) * 0.5 s = 13.2 meters.
Now, let's add up the total distance traveled: 56.78 meters + 13.2 meters = 69.98 meters!
So, it looks like our car moved approximately 69.98 meters from the time the driver saw the deer. That's quite a distance, but luckily, the deer got out of the way just in time. Phew!
I hope I was able to help you with this physics problem while also bringing a smile to your face. If you have any more questions or need further hilarity, feel free to ask!
0 = Vo - a t ignore that line, final speed is not zero and is not needed
To determine how far the car moves in the 2.5 seconds starting from when the driver sees the deer, we can use Newton's laws of motion.
Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The equation that represents this relationship is F = ma, where F is the net force, m is the mass, and a is the acceleration.
In this case, the net force acting on the car is given as 2400 N, and the mass of the car is 1200 kg. We can rearrange the equation to solve for acceleration: a = F / m.
a = 2400 N / 1200 kg
a = 2 m/s²
The car experiences this acceleration for a duration of 2.0 seconds when the driver applies the brakes. Since the car initially moves with a velocity of 95 km/h, it is necessary to convert this velocity to meters per second (m/s).
Given that 1 km = 1000 m and 1 hour = 3600 seconds, we can convert the initial velocity as follows:
95 km/h × (1000 m/1 km) × (1 h/3600 s)
= (95 × 1000) / (3600)
≈ 26.39 m/s
Using the kinematic equation s = ut + (1/2)at², we can now calculate the distance the car moves during the 2.0 seconds of braking:
s = (26.39 m/s)(2.0 s) + (1/2)(2 m/s²)(2.0 s)²
s = 52.78 m + (1/2)(2 m/s²)(4.0 s²)
s = 52.78 m + 4 m
s = 56.78 m
So, the car moves approximately 56.78 meters when the driver applies the brakes for 2.0 seconds.
After the deer jumps out of the way, the driver lifts his foot off the brake pedal, causing the net force to return to zero. Therefore, the car will continue to move at a constant velocity. To find the distance traveled during this time, we need to multiply the final velocity by the time:
s = 26.39 m/s × 0.5 s
s = 13.195 m
Therefore, the car moves approximately 13.195 meters during the 0.5-second period after the driver lifts his foot off the brake pedal.
To find the total distance the car moves in the 2.5 seconds starting from when the driver sees the deer, we simply add the distances calculated from the two periods:
Total distance = 56.78 m + 13.195 m
Total distance = 69.975 m
Hence, the car moves approximately 69.975 meters in the 2.5 seconds starting from when the driver sees the deer.