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Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. What is her final velocity?

Lisa Ford accelerates from 12 m/s to 26 m/s at a rate of 4.2 m/s^2. Over what distance does this acceleration occur?

Ed Foot is traveling at 38.2 m/s when he spots the state police. He decelerates at 8.6 m/s^2 for 2.1 s. What distance does he travel during this time?

What is the acceleration of a car that brakes from 24.2 m/s to 11.9 m/s in 2.85 seconds?

Respuesta :

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

                v = u + at

                   = 6.8 + 4.5 x 3.2

                   = 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

                          v² = u² +2aS

                         S = (v² - u²) / 2a

                            = (26² -12²) / 2 x 4.2

                            = 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

                        s = ut + 1/2 at²

                           =38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

                           = 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

                         v = u + at

∴                        a = (v - u) / t

                            = (11.9 - 24.2) / 2.85

                            = -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

The final velocity, distance and acceleration for each problem are;

1) v = 21.2 m/s

2) S = 63.33 m

3) S = 61.257 m

4) a = -4.32 m/s²

What is the acceleration and distance?

1) We are given;

Initial velocity; u = 6.8 m/s

Acceleration; a = 4.5 m/s²

Time of travel; t = 3.2 s

From Newton's first equation of motion, the final velocity is;

v = u + at

v = 6.8 + (4.5 * 3.2)

v = 21.2 m/s

2) We are given;

Initial velocity; u = 12 m/s

Final velocity; v = 26 m/s

Acceleration; a = 4.2 m/s²

From newton's 3rd equation of motion, the distance travelled is;

S = (v² - u²)/2a

S = (26² - 12²)/(2 * 4.2)

S = 63.33 m

3) We are given;

Initial velocity; u = 38.2 m/s

Deceleration; a = - 8.6 m/s²

Time of travel; t = 2.1 s

From Newton's second equation of motion, the distance is;

S = ut + 0.5at²

S = (38.2 * 2.1) + (0.5 * (-8.6) * 2.1²)

S = 61.257 m

4) We are given;

Initial velocity; u = 24.2 m/s

Final velocity; v = 11.9 m/s

Time taken; t = 2.85 s

From Newton's first equation of motion, we can find the acceleration as;

a = (v - u)/t

a = (11.9 - 24.2)/2.85

a = -4.32 m/s²

Read more about acceleration and distance at; https://brainly.com/question/4931057

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