Physics, 25.09.2020 03:01 mismhan01

A car is traveling along a straight road at a velocity of +36.0 m/s when its engine cuts out. For the next ten seconds the car slows down, and its average acceleration is a1. For the next five seconds the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the fifteen-second period is +24.5 m/s. The ratio of the average acceleration values is a1/a2 = 1.74. Find the velocity of the car at the end of the initial ten-second interval.

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Answer from: lashondrascott

V = 27.07 m/s

Explanation:

Given that a car is traveling along a straight road at a velocity of +36.0 m/s when its engine cuts out. And For the next ten seconds the car slows down, and its average acceleration is a1.

To calculate the acceleration, let use the 1st equation of motion.

V = U - at

As the car is going to rest, V = 0 and a = negative.

V = 36 - a1 × 10

V = 36 - 10a1 (1)

- 10a1 = V - 36

a1 = (36 - V) / 10

Given that For the next five seconds the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the fifteen-second period is +24.5 m/s.

Using the same formula again

V = U - at

24.5 = 36 - 10a1 - a2 × 5

24.5 - 36 = - 10a1 - 5a2

- 11.5 = -( 10a1 + 5a2)

11.5 = 10a1 + 5a2 (2)

But The ratio of the average acceleration values is a1/a2 = 1.74.

a1 = 1.74a2 (3)

Substitute a1 in equation 2

11.5 = 10( 1.74a2) + 5a2

11.5 = 17.4a2 + 5a2

11.5 = 22.4a2

a2 = 11.5/22.4

a2 = 0.5134 m/s^2

Substitutes a2 in equation 3

a1 = 1.74 × 0.5134

a1 = 0.893 m/s^2

Substitutes a1 into equation 1

V = 36 - 10(0.893)

V = 36 - 8.93

V = 27.07 m/s

Therefore, the velocity of the car at the end of the initial ten-second interval is 27.07 m/s

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

Answer from: Quest

access to city

explanation:

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A car is traveling along a straight road at a velocity of +36.0 m/s when its engine cuts out. For th...

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