Question

a car starting from rest moves with constant acceleration of 2.0 m/s2 for 10 s, then travels with constant speed for another 10 s, and then finally slows to a stop with constant acceleration of -2.0 m/s2. how far does it travel? 200 m 300 m 400 m 500 m question 3 (1 point) suppose a ball is thrown straight up, reaches a maximum height, then falls to its initial height. make a statement about the direction of the velocity and acceleration as the ball is going up. both its velocity and its acceleration point upward. its velocity points upward and its acceleration points downward. its velocity points downward and its acceleration points upward. both its velocity and its acceleration points downward. question 4 (1 point) a runner ran the marathon (approximately 42.0 km) in 2 hours and 57 min. what is the average speed of the runner in m/s? 14.2 × 10^3 m/s 124 m/s 3.95 m/s 14.2 m/s

Explanation:

Question 2

Step1: Calculate distance during acceleration

The car starts from rest (\(u = 0\)), acceleration \(a = 2.0\ m/s^2\), time \(t_1 = 10\ s\). Using the formula \(s_1 = ut + \frac{1}{2}at^2\), we substitute \(u = 0\):
\(s_1=\frac{1}{2}\times2\times10^2 = 100\ m\)
Also, find the final velocity \(v\) after acceleration: \(v = u + at = 0 + 2\times10 = 20\ m/s\)

Step2: Calculate distance during constant speed

The speed is constant (\(v = 20\ m/s\)) for \(t_2 = 10\ s\). Using \(s = vt\):
\(s_2 = 20\times10 = 200\ m\)

Step3: Calculate distance during deceleration

The car decelerates from \(v = 20\ m/s\) to \(v = 0\) with \(a=-2.0\ m/s^2\). Using \(v^2 - u^2 = 2as\) (here \(u = 20\), \(v = 0\), \(a=-2\)):
\(0 - 20^2 = 2\times(-2)\times s_3\)
\(-400=-4s_3\)
\(s_3 = 100\ m\)

Step4: Total distance

Total distance \(s = s_1 + s_2 + s_3 = 100 + 200 + 100 = 400\ m\)

When a ball is thrown up, its velocity is in the direction of motion (upward). The acceleration is due to gravity, which acts downward (towards the Earth's center) throughout the motion, including when going up.

Step1: Convert distance to meters and time to seconds

Distance \(d = 42.0\ km = 42000\ m\)
Time \(t = 2\ hours + 57\ min = 2\times3600 + 57\times60 = 7200 + 3420 = 10620\ s\)

Step2: Calculate average speed

Average speed \(v=\frac{d}{t}=\frac{42000}{10620}\approx3.95\ m/s\)

Answer:

400 m

Question 3