A car starts from rest and accelerates for 3.9 s with an acceleration of 5 m/s2 .

How far does it travel? Answer in units of m

d = 0.5a*t^2 = 2.5*3.9^2 = 38 m.

1.28

To find the distance traveled by the car, we can use the equation:

distance = (initial velocity * time) + (1/2 * acceleration * time^2)

Given that the car starts from rest (initial velocity = 0), the equation simplifies to:

distance = 1/2 * acceleration * time^2

Plugging in the values:

distance = 1/2 * 5 m/s^2 * (3.9 s)^2

Calculating:

distance = 1/2 * 5 m/s^2 * 15.21 s^2

distance ≈ 37.945 m

Therefore, the car travels approximately 37.945 meters.

To find the distance traveled by the car, we can use the equation:

d = v₀t + (1/2)at²

Where:
- d is the distance traveled,
- v₀ is the initial velocity (which is 0 since the car starts from rest),
- t is the time taken,
- a is the acceleration.

Plugging in the known values into the equation:

d = (0)(3.9) + (1/2)(5)(3.9)²

First, let's calculate the squared term:

d = (0)(3.9) + (1/2)(5)(15.21)
d = 0 + (1/2)(5)(15.21)
d = 0 + (1/2)(76.05)
d = 0 + 38.025
d = 38.025 m

Therefore, the car travels a distance of 38.025 meters.