All categories

3 years ago

How are velocity and acceleration related?

Physics

2 answers:

Arturiano [62]3 years ago

5 0

Acceleration is the rate of change of velocity

irina1246 [14]3 years ago

5 0

Answer:

<h3>Acceleration is the rate of change of velocity. (when velocity changes -> acceleration exists) If an object is changing its velocity, i.e. changing its speed or changing its direction, then it is said to be accelerating. Acceleration = Velocity / Time (Acceleration)</h3><h3 />

You might be interested in

For which type(s) of signals does noise alter the original information?

nika2105 [10]

Neither analog nor digital

8 0

3 years ago

Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour

Serhud [2]

For the answer to the question above, first find out the gradient.

m = rise/run 
=(y2-y1)/(x2-x1) 

the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" 
(x1,y1) = (3,53) 
(x2,y2) = (6,62) 

sub values back into the equation 
m = (62-53)/(6-3) 
m = 9/3 
m = 3 

now we use a point-slope form to find the the standard form 
y-y1 = m(x-x1) 
where x1 and y1 are any set of point given 
y-53 = 3(x-3) 
y-53 = 3x - 9 
y = 3x - 9 + 53 
y = 3x + 44 

y is the velocity of the car, x is the time.
I hope this helps.

4 0

3 years ago

A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu

vredina [299]

Answer:

a) $v_1=5.5172\ m.s^{-1}$

b) $v_2=-1.5152\ m.s^{-1}$

c) $v_3=4.6154\ m.s^{-1}$

d) $v_{avg}=3.8462\ m.s^{-1}$

Explanation:

Given:

distance down the field in the first interval, $d_1=16\ m$

time duration of the first interval, $t_1=2.9\ s$

distance down the field in the second interval, $d_2=-2.5\ m$

time duration of the second interval, $t_2=1.65\ s$

distance down the field in the third interval, $d_3=24\ m$

time duration of the third interval, $t_3=5.2\ s$

velocity in the first interval:

$v_1=\frac{d_1}{t_1}$

$v_1=\frac{16}{2.9}$

$v_1=5.5172\ m.s^{-1}$

velocity in the second interval:

$v_2=\frac{d_2}{t_2}$

$v_2=\frac{-2.5}{1.65}$

$v_2=-1.5152\ m.s^{-1}$

velocity in the third interval:

$v_3=\frac{d_3}{t_3}$

$v_3=\frac{24}{5.2}$

$v_3=4.6154\ m.s^{-1}$

We know that the average velocity is given as the total displacement per unit time.

$v_{avg}=\frac{d_1+d_2+d_3}{t_1+t_2+t_3}$

$v_{avg}=\frac{16-2.5+24}{2.9+1.65+5.2}$

$v_{avg}=3.8462\ m.s^{-1}$

3 0

3 years ago

Which of the following is a sign of puberty?

umka2103 [35]

I would say D. all of the above

8 0

3 years ago

Read 2 more answers

Help for physics

12345 [234]

Answer:

A. 5 m/s

Explanation:

From the graph, for the first 2 seconds, the graph is a straight line meaning that the slope is a constant.

Average speed of an object is the rate of change of position. Here, the position of the object changes from 0 m to 10 m for a time interval of 2 seconds.

The change in position ( $\Delta x$ ) and time interval ( $\Delta t$ ) are given as:

$\Delta x=10-0=10\ m\\\Delta t=2-0=2\ s$

Therefore, the average speed ( $s_{avg}$ ) is given as the ratio of the total change in position and the time interval for the change.

$s_{avg}=\frac{\Delta x}{\Delta t}=\frac{10-0}{2-0}=\frac{10}{2}=5\ m/s$

Hence, the average speed is 5 m/s.

8 0

3 years ago

Read 2 more answers