A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds. By how much would the force have been reduced if the car had had a crumple zone that increased the collision time to 2.2 seconds? Give your answer to the nearest whole number.

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

[tex]F\Delta t = m\Delta v[/tex]

where:

F is the force exerted on the car

[tex]\Delta t[/tex] is the duration of the collision

m = 1400 kg is the mass of the car

[tex]\Delta v=-27 m/s[/tex] is the change in velocity of the car

We can re-write the equation as

[tex]F=\frac{m\Delta v}{\Delta t}[/tex]

In the 1st collision, the time is 1.5 seconds, so the force is

[tex]F_1=\frac{(1400)(-27)}{1.5}=-25,200 N[/tex]

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

[tex]F_2=\frac{(1400)(-27)}{2.2}=-17,182 N[/tex]

Therefore, the force has been reduced by:

[tex]F_2-F_1=-17,182-(-25,200)=8018 N[/tex]

boffeemadrid boffeemadrid · Answer 2 · 2021-11-22T10:00:08+01:00

The amount of force which is reduced is required.

The force would be reduced by 8018.18 N.

m = Mass of car = 1400 kg

u = Initial velocity = 27 m/s

v = Final velocity = 0 m/s

[tex]\Delta t_1[/tex] = Initial time = 1.5 s

[tex]\Delta t_2[/tex] = Final time = 2.2 s

Force is given by

[tex]F=m\dfrac{v-u}{\Delta t}[/tex]

The difference in force will be

[tex]\Delta F=\left|m(v-u)\left(\dfrac{1}{\Delta t_1}-\dfrac{1}{\Delta t_2}\right)\right|\\\Rightarrow \Delta F=\left|1400(0-27)\left(\dfrac{1}{1.5}-\dfrac{1}{2.2}\right)\right|\\\Rightarrow \Delta F=8018.18\ \text{N}[/tex]

The force would be reduced by 8018.18 N.

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