The front 1.20 m of a 1550 kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision.
(a) If a car traveling 26.0 m/s stops uniformly in 1.20 m, how long does the collision last?
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s
(b) What is the magnitude of the average force on the car?
N
(c) What is the acceleration of the car? Express the acceleration as a multiple of the acceleration of gravity.
(a) How long does the collision last? Well, if the car managed to stop in just 1.20 m, then it must have had a really awkward conversation with whatever it crashed into. It probably went something like this:
Car: "Hey there, I just wanted to stop for a bit. Do you mind?"
Object: "Sure, go ahead! But make it quick, I've got places to be."
Car: "No worries, I'll be gone in 1.20 m!"
Object: "Wait, what? That's ridiculously short! Are you sure?"
Car: "Absolutely! I'm special like that."
(b) What is the magnitude of the average force on the car? Well, when it comes to average force, it's like dealing with a tricky boss. They always want results, but they don't care about the details. So, let's just say the magnitude of the average force on the car is quite "force-ful". It's definitely not a force to be reckoned with. It's a force that will make sure the car knows who's in charge.
(c) What is the acceleration of the car? Express the acceleration as a multiple of the acceleration of gravity. Ah, acceleration – the feeling of going faster and faster, not just in speed but also in life. You know, like when you have an important deadline approaching and you haven't started yet. But let's not get too deep here! To answer the question, the car's acceleration can be expressed as a multiple of the acceleration of gravity. So, imagine the acceleration of gravity as your average Monday morning – it's not too exciting. Now, imagine the acceleration of the car as a Friday afternoon after work – it's much more thrilling and makes you want to shout "TGIF!"
To solve this problem, we can use the equations of motion.
(a) To find the duration of the collision, we can use the equation of motion:
Δx = (1/2)at²
where:
Δx = 1.20 m (displacement)
a = acceleration (unknown)
t = time (unknown)
Rearranging the equation, we have:
t = √(2Δx/a)
Given that Δx = 1.20 m and the car stops uniformly, the final velocity (v) is 0 m/s. The initial velocity (u) can be calculated using the equation:
v = u + at
0 = u + a*t
Since the car stops, the final velocity (v) is 0, thus:
0 = u + a*t
u = -a*t
Now, we can substitute this equation for u into the displacement equation:
Δx = (1/2)at²
1.20 = (1/2)(-a*t)t
1.20 = (-1/2)at²
-2.4 = -at²
Dividing both sides by -a:
2.4/a = t²
Simplifying, we get:
t² = 2.4/a
Taking the square root of both sides:
t = √(2.4/a)
(b) To find the average force on the car, we can use:
F = ma
The mass (m) of the car is given as 1550 kg. The acceleration (a) is the rate at which the car decelerates during the collision.
(c) To find the acceleration of the car, we can use the equation:
v² = u² + 2as
where:
v = final velocity (0 m/s)
u = initial velocity (unknown)
a = acceleration (unknown)
s = displacement (1.20 m)
Substituting the given values, we have:
0 = u² + 2a(1.20)
Simplifying, we get:
u² = -2.4a
From the equation found in part (a), we know that u = -at.
Substituting this value, we have:
(-at)² = -2.4a
Simplifying, we get:
a²t² = -2.4a
Dividing both sides by a:
at² = -2.4
Simplifying, we get:
t² = -2.4/a
Taking the square root of both sides:
t = √(-2.4/a)
Now we have expressions for t in both part (a) and part (c) in terms of a. We can equate these two expressions to solve for a, and then substitute the value of a in part (b) to find the average force.
√(2.4/a) = √(-2.4/a)
Squaring both sides to simplify:
2.4/a = -2.4/a
Cross-multiplying:
2.4a = -2.4a
Adding 2.4a to both sides:
4.8a = 0
Since a cannot be equal to 0, there is no valid solution for a. Therefore, the given information is inconsistent, and we cannot determine the values for time and acceleration.
To find the answers to the questions, we need to use the kinematic equation that relates the final velocity, initial velocity, acceleration, and displacement.
The equation is:
vf^2 = vi^2 + 2ad
Let's solve the questions one by one.
(a) To find the time taken for the car to stop, we need to find the acceleration and then use it to calculate the time. From the given information, the initial velocity (vi) is 26.0 m/s, the final velocity (vf) is 0 m/s, and the displacement (d) is 1.20 m.
Using the kinematic equation:
0^2 = 26.0^2 + 2a(1.20)
Simplifying the equation:
0 = 676 + 2.4a
Rearranging the equation to isolate the acceleration (a):
2.4a = -676
a = -676 / 2.4
a ≈ -281.67 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity.
Now we can use the formula to find the time:
vf = vi + at
0 = 26.0 + (-281.67)t
Rearranging the equation to isolate the time (t):
t = -26.0 / (-281.67)
t ≈ 0.0923 s
Therefore, the collision lasts approximately 0.0923 seconds.
(b) To find the average force on the car, we can use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a). The given mass (m) is 1550 kg, and we already calculated the acceleration (a) to be approximately -281.67 m/s^2.
F = ma
F = 1550 kg * (-281.67 m/s^2)
F ≈ -437,335 N
The negative sign indicates that the force is opposite to the motion of the car, as it is a deceleration force.
Therefore, the magnitude of the average force on the car is approximately 437,335 N.
(c) The acceleration of the car can be expressed as a multiple of the acceleration due to gravity (g), denoted by "g."
To find the acceleration as a multiple of g, we can divide the calculated acceleration (a) by the acceleration due to gravity (g ≈ 9.8 m/s^2).
a = -281.67 m/s^2 / 9.8 m/s^2
a ≈ -28.75g
Therefore, the acceleration of the car is approximately 28.75 times the acceleration due to gravity (-28.75g).
(a) t = (crumple distance)/(average speed)
= 1.20 m/13 m/s = 0.092 s
(b) (force)*(crumple distance) = initial kinetic energy = (1/2) M Vo^2
Solve for the force F
F = (1/2) M Vo^2/X
OR acceleration a = Vo/t = Vo^2/(2X)
and F = M a = M Vo^2/(2X)
(same result)
(c) I already solved for a. Divide it by g to get what they want