Batman was driving the Batmobile at 90 mph (=132 ft/sec), when he sees a brick wall directly ahead. When the Batmobile is 400 feet from the wall, he slams on the brakes, decelerating at a constant rate of 22ft/sec2. Does he stop before he hits the brick wall? If so, how many feet to spare? If not, what is his impact speed? Now the Joker had been driving next to him, also at 90 mph. But the Joker did not hit his brakes as soon as Batman, continuing for 1 second longer than Batman before hitting his brakes, decelerating at a constant rate of 22ft/sec2. How fast is he going when he hits the wall? (Don't worry about Joker - he jettisoned at the last instant, to fight another day!)
Oh boy, looks like Batman and the Joker are up to their usual shenanigans! Let's crunch some numbers and see what happens.
First, let's focus on Batman. He's going at 90 mph, which converts to 132 ft/sec. When Batman slams on the brakes at 400 feet away from the wall, he starts decelerating at 22 ft/sec².
To figure out if Batman stops before hitting the wall, we need to find out how much distance he needs to bring the Batmobile to a halt. We can use the equation:
v² = u² + 2as,
where v is the final velocity (0 ft/sec since he stops), u is the initial velocity (132 ft/sec), a is the deceleration (-22 ft/sec²), and s is the distance.
Rearranging the equation, we get:
s = (v² - u²) / (2a).
Plugging in the values, we get:
s = (0² - 132²) / (2 * -22) = 108 ft.
So Batman stops with 108 feet to spare before slamming into the brick wall. Phew, that was a close call!
Now let's move on to the Joker. He's also driving at 90 mph (132 ft/sec), but he hits his brakes 1 second after Batman. This means the Joker has traveled an additional 132 feet before decelerating.
Using the same equation and values as before, we can calculate the Joker's impact speed. The distance traveled will be 400 + 132 = 532 feet.
s = (v² - u²) / (2a),
s = (0² - 132²) / (2 * -22),
s = 532 ft.
The Joker's impact speed is 132 ft/sec. Well, at least he manages to spare himself by jettisoning! Looks like Batman had the last laugh this time.
Remember, these calculations assume constant deceleration, so don't try them out in the real world. Leave the superhero stuff to Batman and the Joker!
To determine whether Batman stops before hitting the brick wall, we need to calculate the distance he will travel before coming to a stop.
First, we'll find the time it takes for Batman to stop:
Using the formula: v = u + a*t
where v is the final velocity, u is the initial velocity, a is acceleration, and t is time.
Given:
Initial velocity (u) = 132 ft/sec (90 mph)
Acceleration (a) = -22 ft/sec^2 (negative because it's decelerating)
Final velocity (v) = 0 ft/sec (comes to a stop)
0 = 132 + (-22)*t
t = 6 seconds
Next, we'll find the distance Batman will travel before stopping:
Using the formula: s = u*t + 0.5*a*t^2
where s is the distance traveled.
s = 132*6 + 0.5*(-22)*(6^2)
s = 792 - 0.5*22*36
s = 792 - 396
s = 396 feet
Since Batman travels 396 feet before stopping and the wall is 400 feet away when he slams the brakes, he stops just 4 feet before hitting the brick wall.
Now let's calculate the impact speed of the Joker:
The Joker continues driving for 1 second longer than Batman, so his total time of deceleration will be 7 seconds.
Using the same formula to calculate distance:
s = u*t + 0.5*a*t^2
s = 132*7 + 0.5*(-22)*(7^2)
s = 924 - 0.5*22*49
s = 924 - 539
s = 385 feet
Therefore, when the Joker hits the wall, his impact speed would be 90 mph (132 ft/sec) - the same as Batman's initial speed.
To solve this problem, we need to use the kinematic equation of motion. We'll start by calculating how long it will take for Batman to stop.
1. First, let's convert Batman's initial speed of 90 mph to feet per second. Since 1 mile = 5280 feet and 1 hour = 3600 seconds, his initial speed is 90 * 5280 / 3600 = 132 feet per second.
2. Batman's deceleration rate is given as 22 ft/sec^2.
3. We can use the equation v^2 = u^2 + 2as, where v is the final velocity and u is the initial velocity, a is the acceleration, and s is the distance covered.
4. Considering that Batman starts from a distance of 400 feet, his initial velocity is 132 ft/sec, his final velocity is 0 ft/sec, and his acceleration is -22 ft/sec^2 (negative because it is deceleration), we can rearrange the equation to solve for the distance s:
0^2 = 132^2 + 2 * (-22) * s
Simplifying, we get 0 = 17424 - 44s
Rearranging again, we have s = 17424 / 44 = 396 feet.
5. Hence, Batman stops 4 feet before hitting the brick wall, with 396 - 400 = -4 feet to spare.
Now let's calculate the impact speed of the Joker.
1. The Joker starts 1 second later than Batman, but we can assume he accelerates at the same rate as Batman.
2. Since the Joker starts 1 second later, he has covered a distance of 132 * 1 = 132 feet before hitting the brakes.
3. From this point, we can use the same equation as above to calculate the impact speed of the Joker.
a. Initial velocity (u) = 132 ft/sec
b. Final velocity (v) = ?
c. Distance (s) = 400 - 132 = 268 feet
d. Acceleration (a) = -22 ft/sec^2
Using the equation v^2 = u^2 + 2as, we have:
v^2 = 132^2 + 2 * (-22) * 268
Solving the equation, we find v ≈ 100.97 ft/sec.
Hence, the Joker's impact speed, if he had not jettisoned, would be approximately 100.97 ft/sec when he hits the wall.
distance to stop
Vf^2=Vi^2 + 2ad solve for d.
d^2x/dt^2 = -22
dx/dt = 132 - 22 t
x = Xo + 132 t - (1/2)(22) t^2
so x to stop from 132 ft/s
0 = 132 - 22 t
t = 6 seconds to get dx/dt = 0
then
x = 132 (6) - 11 (36)
x = 792 - 396 = 396 ft to stop
so 4 feet to spare
Now I think you can handle Robin';s unfortunate experience.