The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
Answer:
s = 20 m
Explanation:
given,
mass of the roller blader = 60 Kg
length = 10 m
inclines at = 30°
coefficient of friction = 0.25
using conservation of energy
u = 9.89 m/s
Using second law of motion
ma =μ mg
a = μ g
a = 0.25 x 9.8
a = 2.45 m/s²
Using third equation of motion ,
v² - u² = 2 a s
0² - 9.89² = 2 x 2.45 x s
s = 20 m
the distance moved before stopping is 20 m
Answer:
+7.0 m/s
Explanation:
Let's take rightward as positive direction.
So in this problem we have:
a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)
t = 4 s time interval
v = -3.0 m/s is the final velocity (negative because it is leftward)
We can use the following equation:
v = u + at
Where u is the initial velocity
We want to find u, so if we rearrange the equation we find:

and the positive sign means the initial direction was rightward.
Answer:
Initial velocity describes how fast an object travels when gravity first applies force on the object. On the other hand, the final velocity is a vector quantity that measures the speed and direction of a moving body after it has reached its maximum acceleration.
Explanation:
Answer:
a) 
b) 
Explanation:
Given:
- upward acceleration of the helicopter,

- time after the takeoff after which the engine is shut off,

a)
<u>Maximum height reached by the helicopter:</u>
using the equation of motion,

where:
u = initial velocity of the helicopter = 0 (took-off from ground)
t = time of observation


b)
- time after which Austin Powers deploys parachute(time of free fall),

- acceleration after deploying the parachute,

<u>height fallen freely by Austin:</u>

where:
initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)
time of free fall


<u>Velocity just before opening the parachute:</u>



<u>Time taken by the helicopter to fall:</u>

where:
initial velocity of the helicopter just before it begins falling freely = 0
time taken by the helicopter to fall on ground
height from where it falls = 250 m
now,


From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.
<u>remaining time,</u>



<u>Now the height fallen in the remaining time using parachute:</u>



<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>


