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Sagot :
Answer:
1. A.) The moment of the truck before it hits the brakes is 214,400 kg·m/s
B.) The force it takes to stop the truck is approximately 17,290.4 N
Explanation:
1. A.) The given parameters are;
The mass of the truck, m = 8,000 kg
The velocity of the truck when it hits the brakes, u = 26.8 m/s
Momentum = Mass × Velocity
The moment of the truck = The mass of the truck × The velocity of the truck
Therefore;
The moment of the truck before it hits the brakes = 8,000 kg × 26.8 m/s = 214,400 kg·m/s
B.) The amount of momentum lost when the truck comes to a stop = The initial momentum of the truck
The time it takes the truck to come to a complete stop, t = 12.4 s
The deceleration, "a" of the truck is given by the following kinematic equation of motion
v = u - a·t
Where;
v = The final velocity of the truck = 0 m/s
u = The initial velocity = 26.8 m/s
a = the deceleration of the truck
t = The time of deceleration of the truck = 12.4 s
Substituting the known values gives;
0 = 26.8 - a × 12.4
Therefore;
26.8 = a × 12.4
a = 26.8/12.4 ≈ 2.1613
The deceleration (negative acceleration) of the truck, a ≈ 2.1613 m/s²
Force = Mass × Acceleration
The force required to stop the truck = The mass pf the truck × The deceleration (negative acceleration) given to the truck
∴ The force it takes to stop the truck = 8,000 kg × 2.1613 m/s² ≈ 17,290.4 N.
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