Question

Your car is initially at rest when your hit that gas and the car begins to accelerate at a rate of 2.857 m/s/s. The acceleration lasts for 15.5 s. What is the final speed of the car and how much ground does it cover during this acceleration?

Answer 1

The final speed of the car is 44.3835 m/s, and it travels approximately 345.789375 meters during the acceleration from rest at a rate of 2.857 m/s² over 15.5 seconds.

Step-by-step explanation:

To calculate the final speed of the car and the distance it covers during acceleration from rest, we use two equations of motion:

Final velocity (v) = initial velocity (u) + acceleration (a) × time (t).

Since the car starts from rest, the initial velocity u = 0, and thus the equation simplifies to: v = a × t.

The distance (s) covered can be calculated using the formula: s = u × t + 0.5 × a × t².

Again, as the car starts from rest, u = 0, leading to: s = 0.5 × a × t².

For the car accelerating at 2.857 m/s² for 15.5 seconds, we can calculate:

• Final velocity (v) = 2.857 m/s² * 15.5 s = 44.3835 m/s.
• Distance covered (s) = 0.5 * 2.857 m/s² * (15.5 s)² = 345.789375 m.

Therefore, the final speed of the car is 44.3835 m/s, and it covers a ground of approximately 345.789375 meters during this acceleration.

Answer 2

Vf = 44.56 [m/s]

Step-by-step explanation:

In order to solve this problem we must use the following expression of kinematics.

`v_(f) = v_(i)+(a*t)`

where:

Vf = final velocity [m/s]

Vi = initial velocity = 0

a = acceleration = 2.857 [m/s^2]

t = time = 15.5 [s]

Note: the initial velocity is equal to zero as the car begins its movement from rest, or with an initial velocity equal to zero.

Vf = 0 + (2.875*15.5)

Vf = 44.56 [m/s]