# The object has a speed of 15 m/s. It took the object 5 second to travel, how far did the object go?

The object has a speed of 15 m/s. It took the object 5 second to travel, how far did the object go?

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1. lollollollollol1 says:

1. D) decreases, inverse

The average speed for each lap is given by:

$v=\frac{S}{t}$

where S is the length of the lap (400 m) while t is the time taken to complete the lap. Since the time t increases at each lap, we see from the formula that v, the speed, decreases at each lap. This is an example of inverse relationship: when one of the two quantities increases (the time), the other one decreases (the speed).

2. A) Velocity on the y-axis, time on the x-axis

Acceleration is defined as the change in velocity divided by the time taken:

$a=\frac{\Delta v}{\Delta t}$

In a graph, the slope of a line is given by the increment in y divided by the increment in x:

$s=\frac{\Delta y}{\Delta x}$

Therefore, if we put velocity on the y-axis and time on the x-axis, we immediately see that the acceleration corresponds to the slope of the curve.

3. D)velocity

In fact, velocity consists of 1) speed and 2) direction. The speed can be calculated as:

$v=\frac{S}{t}$

where S is the distance covered and t the time taken. So, by knowing these two quantities, Tim can calculate the speed of the trip. Tim also knows the direction of motion, so he can determine the velocity of the trip.

4. B)equal, different

The difference between speed and velocity is simple: speed is a scalar, and it is just the magnitude of the velocity, so "how fast" is the object moving, while velocity takes also into account the direction of motion, so it is a vector.

Therefore: the two cars have same speed (25 m/s), but they have different velocities, since they are moving into different directions (car A is moving from NY to Miami, while car B is moving from NY to Chicago).

28. B) 11 m

The distance travelled by the bag in free fall is given by:

$S=\frac{1}{2}at^2$

where $a=9.8 m/s^2$ is the acceleration of gravity and t=1.5 s is the time. Replacing these numbers into the formula, we find

$S=\frac{1}{2}(9.8 m/s^2)(1.5 s)^2 =11 m$

29. B)10 m/s south

The boat is moving 20 m/s south, relative to the ground. The passenger is moving 10 m/s north, relative to the boat. Considering south as positive direction, the passenger's velocity relative to the ground is

$v' = v_b + v_p = 20 m/s + (-10 m/s) = 10 m/s$

and the positive sign means it is due south.

30. B)direction

In fact, velocity is a vector, so it consists of a magnitude (the speed) and a direction. The speed can be determined by the distance and the time, in fact speed is defined as

$v=\frac{d}{t}$

where d=distance and t=time; however, in order to determine velocity, we also need to know the direction of motion.

33. B)a velocity.

In fact, velocity is a vector, and it consists of a magnitude (the speed) and a direction. In this case, 10 m/s is the speed, while north is the direction.

34. C)Does the measurement include direction?

As stated in the previous question, a vector includes both magnitude and direction, while a scalar includes only a magnitude. Therefore, by asking if the measurement includes a direction, we are able to determine if the quantity is a vector or not.

2. maelonramirez says:

question one is C)

question two is D)

question three is D)

question four is B)

question five is B)

question six is B)

question seven is B)

question eaghit is C)

hope ths helps pleas leave comments to let me know if I got anything wrong.

3. paul1963 says:

Seventy-Five Metres

Explanation:

You can solve this simply by multiplying 15 m/s by 5 seconds

$15 \frac{m}{s} * 5s\\=75\frac{ms}{s}\\= 75m$

4. ciralove2004 says:

1) The acceleration of the runner is $-3.33 m/s^2$

2) The time taken is 4.38 h

3) The final velocity is 13.7 m/s north

4) Displacement: 10 m forward, distance: 40 m

Explanation:

1)

The acceleration of a body is given by the equation

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

For the runner in this problem, we have

u = 10 m/s

v = 0 (he comes to a stop)

t = 3 s

Substituting, we find

$a=\frac{0-10}{3}=-3.33 m/s^2$

2)

The speed of an object in uniform motion is given by

$v=\frac{d}{t}$

where

d is the distance covered

t is the time taken

For the car in this problem, we have:

d = 350 km (distance covered)

v = 80 km/h (speed)

Solving for t, we find the time required for the trip:

$t=\frac{d}{v}=\frac{350}{80}=4.38 h$

3)

For an object in uniformly accelerated motion, we can use the following suvat equation:

$v=u+at$

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the object in this problem, taking north as positive direction, we have

u = 5.0 m/s (initial velocity)

$a=3.0 m/s^2$ (acceleration)

And substituting t = 2.9 s, we find the final velocity:

$v=5.0 + (3.0)(2.9)=13.7 m/s$ (north)

4)

a) Displacement is a vector connecting the initial position to the final position of motion. Its magnitude is equal to the shortest distance in a straight line between the initial and the  final position.

In this problem, you walk 25 meters forward and then 15 meters backward: therefore, the final position is

x = 25 - 15 = +10 m (forward)

Therefore, the displacement is

$\Delta x = +10 - 0 = +10 m$ (forward)

b) Distance is a scalar quantity measuring the total length of the path covered during the motion, regardless of the direction. In  this case therefore, distance is simply calculated by adding the distance covered in the first and second part of the motion:

$d=25 + 15 = 40 m$