A truck on a straight road starts from rest accelerating at 2.0 m/s2 until it reaches a speed of 20 m/s. Then the truck travels for 20 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.0 s. What is the average velocity of the truck for the motion described?

Add up the distances travelled during the acceleration, constant-velocity, and deceleration phases of the journey. Then divide that total distance travelled by the elapsed time (10 + 20 + 5 s) to get the average velocity. If you need additional help, show your worka and it will be provided.

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If instead, the swimmer makes 25.5 laps in the same pool in 1.5 h, determine the total distance the swimmer traveled, the swimmer's total displacement, the swimmer's average velocity, and the swimmer's average speed.

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To find the average velocity of the truck, we need to calculate the total displacement and the total time taken for the entire motion described.

Let's break down the motion into three parts:

Part 1: Acceleration phase
The truck starts from rest and accelerates at a rate of 2.0 m/s^2 until it reaches a speed of 20 m/s. To find the time taken (t1) and the displacement (s1) during this phase, we can use the following equations of motion:

v = u + at (where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time)
s = ut + (1/2)at^2 (where s is the displacement)

Initially, the truck is at rest (u = 0 m/s), and the final velocity is given as 20 m/s (v = 20 m/s), and acceleration is 2.0 m/s^2 (a = 2.0 m/s^2). We can rearrange the first equation to solve for time:

t1 = (v - u) / a

Substituting the values, we get:
t1 = (20 - 0) / 2.0 = 10 s

Now, using the second equation of motion, we can find the displacement during this period:

s1 = ut1 + (1/2) a t1^2
s1 = 0 + (1/2) * 2.0 * (10)^2
s1 = 100 m

Part 2: Constant speed phase
After reaching a speed of 20 m/s, the truck travels at a constant speed for 20 s. During this phase, the velocity remains constant. So, the displacement (s2) is the product of the velocity and time:

s2 = v * t2
s2 = 20 * 20 = 400 m

Part 3: Braking phase
The truck takes 5.0 s to stop uniformly. During uniform deceleration, the displacement (s3) can be calculated using the equation:

s = vt - (1/2)at^2

Since the truck comes to a stop, the final velocity (v) is 0 m/s, and the acceleration (a) can be determined by using the first equation of motion:

v = u + at
0 = 20 + a * 5.0
a = -4.0 m/s^2 (since acceleration is opposite to the initial velocity)

Now, using the equation for displacement, we have:

s3 = vt3 - (1/2)at3^2
s3 = 0 * 5.0 - (1/2) * (-4.0) * (5.0)^2
s3 = 50 m

To find the total displacement (s_total), we add up the displacements from the three parts:

s_total = s1 + s2 + s3
s_total = 100 + 400 + 50
s_total = 550 m

The total time taken (t_total) is the sum of the times taken for each phase:

t_total = t1 + t2 + t3
t_total = 10 + 20 + 5
t_total = 35 s

Now, we can calculate the average velocity (v_avg) using the formula:

v_avg = s_total / t_total
v_avg = 550 / 35
v_avg = 15.71 m/s

Therefore, the average velocity of the truck for the motion described is approximately 15.71 m/s.