Question

1. a car with an initial speed of 6.58 m/s accelerates at a unif rate of 0.92 m/s² for 3.6 seconds.

a. find the speed of the car after the 3.6 s.
b. calculate the displacement of the car during this time.

Explanation:

Part (a)

Step1: Recall the kinematic equation for final velocity

The kinematic equation for final velocity \( v \) when there is uniform acceleration is \( v = u + at \), where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
Here, \( u = 6.58 \, \text{m/s} \), \( a = 0.92 \, \text{m/s}^2 \), and \( t = 3.6 \, \text{s} \).

Step2: Substitute the values into the equation

Substitute the given values into \( v = u + at \):
\( v = 6.58 + (0.92\times3.6) \)
First, calculate \( 0.92\times3.6 = 3.312 \)
Then, \( v = 6.58 + 3.312 = 9.892 \, \text{m/s} \)

Step1: Recall the kinematic equation for displacement

The kinematic equation for displacement \( s \) with uniform acceleration is \( s = ut + \frac{1}{2}at^2 \), where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
Here, \( u = 6.58 \, \text{m/s} \), \( a = 0.92 \, \text{m/s}^2 \), and \( t = 3.6 \, \text{s} \).

Step2: Substitute the values into the equation

Substitute the values into \( s = ut + \frac{1}{2}at^2 \):
First, calculate \( ut = 6.58\times3.6 = 23.688 \)
Then, calculate \( \frac{1}{2}at^2=\frac{1}{2}\times0.92\times(3.6)^2 \)
First, \( (3.6)^2 = 12.96 \)
Then, \( \frac{1}{2}\times0.92\times12.96 = 0.46\times12.96 = 5.9616 \)
Now, add the two parts: \( s = 23.688 + 5.9616 = 29.6496 \, \text{m} \)

Answer:

The speed of the car after 3.6 s is \( \boldsymbol{9.892 \, \text{m/s}} \)

Part (b)