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A friend claims that as long as he has his seat belt on, he can hold on to a 14 kg child in a 64 mi/h head-on collision with a brick wall in which the car passenger compartment comes to a stop in 0.05 s. Show that the violent force during the collision will tear the child from his arms. A child should always be in a toddler seat secured with a seat belt in the back seat of a car.

2 Answers

8 votes

Final answer:

Calculations based on the scenario indicate an extremely high force of 8064 N would be exerted on a 14 kg child in a head-on collision at 64 mi/h, which would be impossible to withstand by holding the child. It highlights the importance of properly securing a child in a car seat.

Step-by-step explanation:

Force in a Head-On Collision

During a head-on collision with a fixed object, like a brick wall, the deceleration is extremely rapid. Given a 64 mi/h (which is roughly 28.8 m/s) collision and a stopping time of 0.05 s, we can calculate the force experienced by a 14 kg child using Newton's second law of motion (F = ma). First, we convert the velocity to meters per second and then calculate the acceleration (or deceleration in this case) using the formula a = Δv / Δt, which gives us a = 28.8 m/s / 0.05 s. The result is a deceleration of 576 m/s².

Next, we apply the formula F = ma to find the force exerted on the child. With a mass m of 14 kg and the calculated deceleration a, the force would be F = 14 kg * 576 m/s². The force comes out to be 8064 N (newtons), which is a tremendous amount of force and far beyond what a person could hold onto during a crash.

It is essential to understand that in the forces involved in collisions are too great for a person to safely hold on to a child even with a seat belt on. A child must be secured in an appropriate car seat that is designed to safely distribute and absorb the forces of a crash. To rely on holding a child is both dangerous and unrealistic.

User Dan Friedman
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13 votes

Answer:

Force is too large for a person to exert.

Step-by-step explanation:

m = Mass of child = 14 kg

u = Initial velocity of compartment = 64 mi/h =
(64* 1609.34)/(3600)=28.61\ \text{m/s}

v = Final velocity of compartment = 0

t = Time taken by the compartment to stop = 0.05 s

Force is given by


F=ma\\\Rightarrow F=m(v-u)/(t)\\\Rightarrow F=14* (0-28.61)/(0.05)\\\Rightarrow F=-8010.8\ \text{N}

The person has to exert a force of
-8010.8\ \text{N} in the opposite direction to keep holding the child. This is a huge amount of force and cannot be done by a person's physical strength alone.

User Astropringles
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3.7k points