5. a cyclist starts from rest and speeds up to 8 m/s in 4 seconds, then maintains 8 m/s for 6 seconds while…

5. a cyclist starts from rest and speeds up to 8 m/s in 4 seconds, then maintains 8 m/s for 6 seconds while riding on a straight path. reaching a hill, the cyclist slows down uniformly from 8 m/s to 0 m/s in 3 seconds. after stopping, the cyclist rolls backward down the hill, reaching - 4 m/s in 2 seconds, and continues at that speed for 4 seconds. sketch the motion of the cyclist on the graph below.\n6. sally drove at a speed of 50 km/h south for 2 hours. how far did she travel?\n7. a skateboarder with a mass of 60 kg is moving at a velocity of 3 m/s. what is the skateboarder’s momentum?\n8. what is the velocity of a car that is moving with a momentum of 35 kg m/s and has a mass of 945 kg?\n9. a bike rider went from 3 m/s to 8 m/s in 2 seconds, what was his acceleration?\n10. what is the mass of a car with a momentum of 6480 kg x m/s and a velocity of 18 m/s?

5. a cyclist starts from rest and speeds up to 8 m/s in 4 seconds, then maintains 8 m/s for 6 seconds while riding on a straight path. reaching a hill, the cyclist slows down uniformly from 8 m/s to 0 m/s in 3 seconds. after stopping, the cyclist rolls backward down the hill, reaching - 4 m/s in 2 seconds, and continues at that speed for 4 seconds. sketch the motion of the cyclist on the graph below.\n6. sally drove at a speed of 50 km/h south for 2 hours. how far did she travel?\n7. a skateboarder with a mass of 60 kg is moving at a velocity of 3 m/s. what is the skateboarder’s momentum?\n8. what is the velocity of a car that is moving with a momentum of 35 kg m/s and has a mass of 945 kg?\n9. a bike rider went from 3 m/s to 8 m/s in 2 seconds, what was his acceleration?\n10. what is the mass of a car with a momentum of 6480 kg x m/s and a velocity of 18 m/s?

Answer

Question 6

Explanation:

Step1: Recall distance - speed - time formula

The formula for distance $d$ is $d = s\times t$, where $s$ is speed and $t$ is time.

Step2: Substitute values

Given $s = 50$ km/h and $t=2$ h. Then $d=50\times2$.

Answer:

100 km

Question 7

Explanation:

Step1: Recall momentum formula

The formula for momentum $p$ is $p = m\times v$, where $m$ is mass and $v$ is velocity.

Step2: Substitute values

Given $m = 60$ kg and $v = 3$ m/s. Then $p=60\times3$.

Answer:

180 kg m/s

Question 8

Explanation:

Step1: Recall momentum formula and solve for velocity

The formula for momentum is $p = m\times v$. Rearranging for $v$ gives $v=\frac{p}{m}$.

Step2: Substitute values

Given $p = 35$ kg m/s and $m = 945$ kg. Then $v=\frac{35}{945}=\frac{1}{27}\approx0.037$ m/s.

Answer:

$\frac{1}{27}$ m/s (or approximately 0.037 m/s)

Question 9

Explanation:

Step1: Recall acceleration formula

The formula for acceleration $a$ is $a=\frac{v_f - v_i}{t}$, where $v_f$ is final - velocity, $v_i$ is initial - velocity, and $t$ is time.

Step2: Substitute values

Given $v_f = 8$ m/s, $v_i = 3$ m/s, and $t = 2$ s. Then $a=\frac{8 - 3}{2}=\frac{5}{2}=2.5$ m/s².

Answer:

2.5 m/s²

Question 10

Explanation:

Step1: Recall momentum formula and solve for mass

The formula for momentum is $p = m\times v$. Rearranging for $m$ gives $m=\frac{p}{v}$.

Step2: Substitute values

Given $p = 6480$ kg m/s and $v = 18$ m/s. Then $m=\frac{6480}{18}=360$ kg.

Answer:

360 kg