"Caleb misjudges while parking, and bumps his car into a brick wall at 3 MPH. His bumper brings the car to a stop in about 2 inches. We wish to estimate the force the wall exerts on the car during this bump. The force is about:"

Respuesta :

Explanation:

Using equation of motion to determine the acceleration of the car,

vf^2 = vi^2 + 2 * a * S,

vf = 0

0 = vi^2 + 2 * a * S

Converting mph to m/s,

3 mph * 5280 ft/mi * 12 in/ft * 2.54 cm/in * 1 m/100 cm * 1 h/3600 s

= 3 * 0.445

v = 1.335 m/s

Converting in to m,

2 in * 2.54 cm/in * 1 m/100 cm = 0.0254 m

= 2 * 0.0254

S = 0.0508 m

0 = 1.335^2 + 2 * a * 0.0508

a = -1.335^2 ÷ 0.1013

= -17.54 m/s^2

Mass of car (assumed) = 2000 kg

Force = ma

= 2000 × 17.54

= 35.08 kN.

The value of force will be 35.08kN.

From the equation of acceleration,

[tex]v_f^2 = v_i^2 + 2 * a * s[/tex]....................(1)

where

Final velocity, [tex]v_f[/tex]= 0

Initial velocity= [tex]v_i[/tex]

Therefore,

[tex]0 = v_i^2 + 2 * a * s[/tex]

  • Unit conversions for mph to m/s,

[tex]3 mph * 5280 ft/mi * 12 in/ft * 2.54 cm/in * 1 m/100 cm * 1 h/3600 s = 3 * 0.445[/tex]

So the value of velocity will be, v = 1.335 m/s

  • Unit conversion for inches to meters,

[tex]2 in * 2.54 cm/in * 1 m/100 cm = 0.0254 m = 2 * 0.0254[/tex]

So the value of displacement will be, s = 0.0508 m

Now substituting the values in equation (1)

[tex]0 = 1.335^2 + 2 * a * 0.0508\\\\a =\frac{ -1.335^2}{0.1013} \\\\a= -17.54 m/s^2[/tex]

Given:

Mass of car (assumed) = 2000 kg

So, adding values of mass and acceleration in the formula of Force.

[tex]\text{Force} = m*a\\\\\text{Force}= 2000 * 17.54\\\\\text{Force}= 35.08 kN[/tex]

Thus, the value of force will be 35.08kN.

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