Kathy tests her new sports car by racing with stan, an experienced racer. both start from rest, but kathy leaves the starting line 1.00 s after stan does. stan moves with a constant acceleration of 3.1 m/s2 while kathy maintains an acceleration of 4.99 m/s. 2 (a) find the time at which kathy overtakes stan. s from the time kathy started driving (b) find the distance she travels before she catches him (c) find the speeds of both cars at the instant she overtakes him. kathy m/s stan m/s
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Ответ:
(a) t=3.87 s :time at which Kathy overtakes Stan
(b) d=37.36 m
(c) vf₁ = 15.097 m/s : Stan's final speed
vf₂ = 19.31 m/s : Kathy's final speed
Explanation:
kinematic analysis
Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:
vf= v₀+at Formula (1)
vf²=v₀²+2*a*d Formula (2)
d= v₀t+ (1/2)*a*t² Formula (3)
Where:
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Nomenclature
d₁: Stan displacement
t₁ : Stan time
v₀₁: Stan initial speed
vf₁: Stan final speed
a₁: Stan acceleration
d₂: car displacement
t₂ : Kathy time
v₀₂: Kathy initial speed
vf₂: Kathy final speed
a₂: Kathy acceleration
Data
v₀₁ = 0
v₀₂ = 0
a₁ = 3.1 m/s²
a₂= 4.99 m/s²
t₁ = (t₂ +1) s
Problem development
By the time Kathy overtakes Stan, the two will have traveled the same distance:
d₁ = d₂
t₁ = (t₂ +1)
We aplpy the Formula (3)
d₁ = v₀₁t₁ + (1/2)*a₁*t₁²
d₁ = 0 + (1/2)*(3.1)*t₁²
d₁ = 1.55*t₁² ; Stan's cinematic equation 1
d₂ = v₀₂t₂ + (1/2)*a₂*t₂²
d₂ = 0 + (1/2)*(4.99)*t₂²
d₂ = 2.495* t₂² : Kathy's cinematic equation 2
d₁ = d₂
equation 1 = equation 2
1.55*t₁² = 2.495* t₂² , We replace t₁ = (t₂ +1)
1.55* (t₂ +1) ² = 2.495* t₂²
1.55* (t₂² +2t₂+1) = 2.495* t₂²
1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²
1.55t₂²+3.1t₂+1.55=2.495t₂²
(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0
0.905t₂² - 3.1t₂ - 1.55 = 0 Quadratic equation
Solving the quadratic equation we have:
(a) t₂ = 3.87 s : time at which Kathy overtakes Stan
(b) Distance in which Kathy catches Stan
we replace t₂ = 3.87 s in equation 2
d₂ = 2.495*( 3.87)²
d₂ = 37.36 m
(c) Speeds of both cars at the instant Kathy overtakes Stan
We apply the Formula (1)
vf₁= v₀₁+a₁t₁ t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s
vf₁= 0+3.1* 4.87
vf₁ = 15.097 m/s : Stan's final speed
vf₂ = v₀₂+a₂ t₂
vf₂ =0+4.99* 3.87
vf₂ = 19.31 m/s : Kathy's final speed
Ответ:
1) 24500 J (29400 J if we include also the water)
The gravitational potential energy of an object is given by:
where m is the mass, g is the gravitational acceleration and h is the heigth above the ground.
Therefore, the gravitational potential energy of the cylinder is given by:
If we count also the water inside the cylinder, the total mass is 5.0 kg + 1.0 kg = 6.0 kg, and the total gravitational potential energy would be
2) 24530 J
The heat energy added to the water is given by:
where
m = 1.0 kg is the mass of the water
C = 4186 J/kg C is the specific heat of the water
is the increase in temperature
Substituing, we find
3) 1.0 kg
The mass of the water is given, and it is m = 1.0 kg.
4)
The temperature change of the water is the difference between its final temperature and its initial temperature:
5)
The specific heat of the water is