Create a velocity - time diagram as shown below.
The car accelerates from rest to a velocity, v, followed by deceleration to rest at the expected height.
Acceleration phase:
The velocity v is attained in time t₁ with acceleration of 0.6 ft/s², therefore
v = (0.6 ft/s²)(t₁ s) = 0.6t₁ ft/s
t₁ = v/0.6 = 1.6667v s
The distance traveled is
h₁ = (1/2)*(0.6 ft/s²)*(t₁ s)²
= 0.3(1.6667v)²
= 0.8334v² ft
Deceleration phase:
The car comes to rest from an initial velocity of v with a deceleration of 0.3 ft/s² in time t₂. Therefore
v - (0.3 ft/s²)*(t₂ s) = 0
t₂ = v/0.3 = 3.3333v s
The distance traveled is
h₂ = vt₂ - (1/2)*(0.3 ft/s²)*(t₂ s)²
= 3.3333v² - 0.15*(3.3333v)²
= 1.6667v² ft
The total distance traveled should be 48 ft, therefore
0.8334v² + 1.6667v² = 48
2.5v² = 48
v = 4.3818 ft/s (should not exceed 8 ft/s, so it is okay).
The shortest time to make the lift is
t₁ + t₂ = 1.6667v + 3.3333v
= 5v
= 5*4.3818
= 21.9 s
Answer: 21.9 s
Answer:
2.88m/s
Explanation:
Given parameters:
Displacement = 7.2m
Time taken = 2.5s
Unknown:
Velocity of the plane = ?
Solution:
Velocity is the displacement divided by the time taken.
Velocity =
So;
Velocity = = 2.88m/s
The answer is not 'A'.
Using the numbers given in the question, it's 'B'.
It WOULD be 'A' if the number in B were 71 or less.
________________________________
<span>This is not as simple as it looks.
What quantity are we going to compare between the two cases ?
Yes, I know ... the "amount of work". But how to find that from the
numbers given in the question ?
Is it the same as the change in speed ?
Well ? Is it ?
NO. IT's NOT.
In order to reduce the car's speed, the brakes have to absorb
the KINETIC ENERGY, and THAT changes in proportion to
the SQUARE of the speed. ( KE = 1/2 m V² )
Case 'A' :
The car initially has (1/2 m) (100²)
= (1/2m) x 10,000 units of KE.
It slows down to (1/2 m) x (70²)
= (1/2m) x 4,900 units of KE.
The brakes have absorbed (10,000 - 4,900) = 5,100 units of KE.
Case 'B' :
The car initially has (1/2 m) (79²)
= (1/2m) x 6,241 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 6,241 units of KE.
Just as we suspected when we first read the problem,
the brakes do more work in Case-B, bringing the car
to a stop from 79, than they do when slowing the car
from 100 to 70 .
But when we first read the problem and formed that
snap impression, we did it for the wrong reason.
Here, I'll demonstrate:
Change Case-B. Make it "from 71 km/h to a stop".
Here's the new change in kinetic energy for Case-B:
The car initially has (1/2 m) (71²)
= (1/2m) x 5,041 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 5,041 units of KE.
-- To slow from 100 to 70, the brakes absorbed 5,100 units of KE.
-- Then, to slow the whole rest of the way from 71 to a stop,
the brakes absorbed only 5,041 units of KE.
-- The brakes did more work to slow the car the first 30 km/hr
than to slow it to a complete stop from 71 km/hr or less.
That's why you can't just say that the bigger change in speed
requires the greater amount of work.
______________________________________
It works exactly the same in the opposite direction, too.
It takes less energy from the engine to accelerate the car
from rest to 70 km/hr than it takes to accelerate it the
next 30, to 100 km/hr !</span>
(The exact break-even speed for this problem is 50√2 km/h,
or 70.711... km/hr rounded. )
<h3><u>Answer</u>;</h3>
1600 years
<h3><u>Explanation</u>;</h3>
- Half life is the time taken for a radioactive isotope to decay by half of its original amount.
- We can use the formula; N = O × (1/2)^n ; where N is the new mass, O is the original amount and n is the number of half lives.
- A sample of radium-226 takes 3200 years to decay to 1/4 of its original amount.
Therefore;
<em>1/4 = 1 × (1/2)^n</em>
<em>1/4 = (1/2)^n </em>
<em>n = 2 </em>
Thus; <em>3200 years is equivalent to 2 half lives.</em>
<em>Hence, the half life of radium-226 is 1600 years</em>