Question

1. holt sf 02d 02 20 pts possible

part 1 of 2
an automobile with an initial speed of 5.14 m/s accelerates uniformly at the rate of 3.3 m/s².
find the final speed of the car after 6.0 s.
answer in units of m/s. answer in units of m/s.
part 2 of 2
find the displacement of the car after 6.0 s.
answer in units of m. answer in units of m.

Explanation:

Part 1 of 2

Step1: Recall the kinematic equation for final velocity

The formula for final velocity \( v \) in uniformly accelerated motion is \( v = u + at \), where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
Given \( u = 5.14 \, \text{m/s} \), \( a = 3.3 \, \text{m/s}^2 \), and \( t = 6.0 \, \text{s} \).

Step2: Substitute the values into the formula

Substitute \( u = 5.14 \), \( a = 3.3 \), and \( t = 6.0 \) into \( v = u + at \):
\[
v = 5.14 + (3.3 \times 6.0)
\]
First, calculate \( 3.3 \times 6.0 = 19.8 \).
Then, \( v = 5.14 + 19.8 = 24.94 \, \text{m/s} \).

Step1: Recall the kinematic equation for displacement

The formula for displacement \( s \) in uniformly accelerated motion is \( s = ut + \frac{1}{2}at^2 \), where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
Given \( u = 5.14 \, \text{m/s} \), \( a = 3.3 \, \text{m/s}^2 \), and \( t = 6.0 \, \text{s} \).

Step2: Substitute the values into the formula

Substitute \( u = 5.14 \), \( a = 3.3 \), and \( t = 6.0 \) into \( s = ut + \frac{1}{2}at^2 \):
First, calculate \( ut = 5.14 \times 6.0 = 30.84 \).
Then, calculate \( \frac{1}{2}at^2 = \frac{1}{2} \times 3.3 \times (6.0)^2 \).
\( (6.0)^2 = 36 \), so \( \frac{1}{2} \times 3.3 \times 36 = 1.65 \times 36 = 59.4 \).
Now, add the two results: \( s = 30.84 + 59.4 = 90.24 \, \text{m} \).

Answer:

\( 24.94 \, \text{m/s} \)

Part 2 of 2