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jeyben [28]
3 years ago
13

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

nelastic and takes place over an interval of 0.203 s. Assume that no brakes are applied during the collision and the car strikes the rear of the truck. Neglect the friction between the vehicles and the ground.
What is the average x component of the acceleration of the car during the collision?
What is the average x component of the acceleration of the truck during the collision?
Physics
1 answer:
tresset_1 [31]3 years ago
5 0

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

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Answer:

Rate = 1.33 \pm 0.055 beats per second

Explanation:

Number of heart beats = 40 \pm 1

time taken = 30.0 \pm 0.5 s

now we have

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now rate of heart beat is defined as number of heart beat per unit of time

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3 years ago
A child is riding in a child-restraint chair, securely fastened to the seat of a car. Assume the car has speed 47 km/h when it h
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Answer: F = 1235 N

Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: F_{net}=m.a

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