Question

an object starts at 3.39 m/s and accelerates at 7.34 m/s/s for 8.34 sec. how fast will it be going now? an object goes from 2.75 m/s to 23.23 m/s in 4 sec. what is its acceleration?

Explanation:

Problem 4:

Step1: Recall the kinematic formula for final velocity

The formula for final velocity \( v \) when an object starts with initial velocity \( u \), accelerates with acceleration \( a \) for time \( t \) is \( v = u + at \).
Given \( u = 3.39 \, \text{m/s} \), \( a = 7.34 \, \text{m/s}^2 \), \( t = 8.34 \, \text{sec} \).

Step2: Substitute the values into the formula

\( v = 3.39 + (7.34\times8.34) \)
First, calculate \( 7.34\times8.34 \): \( 7.34\times8.34 = 61.2156 \)
Then, \( v = 3.39 + 61.2156 = 64.6056 \, \text{m/s} \)

Step1: Recall the formula for acceleration

Acceleration \( a \) is given by the change in velocity divided by time, i.e., \( a=\frac{v - u}{t} \), where \( v \) is final velocity, \( u \) is initial velocity, and \( t \) is time.
Given \( u = 2.75 \, \text{m/s} \), \( v = 23.23 \, \text{m/s} \), \( t = 4 \, \text{sec} \).

Step2: Substitute the values into the formula

\( a=\frac{23.23 - 2.75}{4} \)
First, calculate the numerator: \( 23.23 - 2.75 = 20.48 \)
Then, \( a=\frac{20.48}{4}=5.12 \, \text{m/s}^2 \)

Answer:

The final velocity of the object is \( 64.61 \, \text{m/s} \) (rounded to two decimal places)

Problem 5: