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Elenna [48]
3 years ago
10

The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c

ar traveling 24.0 m/s stops uniformly in 1.20 m, how long does the collision last? s (b) What is the magnitude of the average force on the car? N (c) What is the magnitude of the acceleration of the car? Express the acceleration as a multiple of the acceleration of gravity. g
Physics
1 answer:
eimsori [14]3 years ago
3 0

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

a = 53.3g

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