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.