<h2>Answer </h2>
C. 40 m/s
<h2>Explanation </h2>
To solve this, we are using the kinematic equation for final velocity:
where
is the final velocity
is the initial velocity
is the acceleration
is the time
We know from our a problem that a motorcycle moving at 20 m/s accelerates uniformly 2 m/s^2 for 10 seconds, so , , and .
Now we can replace the values
We can conclude that the velocity of the motorcycle will be 40 m/s after 10 seconds.
To solve the problem it is necessary to apply the concepts related to Chvorinov's Law, which states that
Where,
= Volume cube
= Superficial Area from Cube
= Volume Rectangle
= Superficial Area from Rectangle
Our values are given as (I will try to develop the problem in English units for ease of calculations),
Applying the Chvorinov equation we have to,
The stipulated time for the cube is 14.5 then,
Answer:
0.56 m/s
Explanation:
The speed of the head at the end of the interval in each case is the area under the acceleration curve. Then the difference in speeds is the difference in areas.
We can find the area geometrically, using formulas for the area of a triangle and of a trapezoid.
A = 1/2bh . . . . area of a triangle
A = 1/2(b1 +b2)h . . . . area of a trapezoid
If h(t) is the acceleration at time t for a helmeted head, the area under that curve will be (in units of mm/s) ...
Vh = 1/2(h(3)·3) +1/2(h(3) +h(4))·1 +1/2(h(4) +h(6))·2 +1/2(h(6))·1
Vh = 1/2(4h(3) +3h(4) +3h(6)) = 1/2(4·40 +3·40 +3·80) = 260 . . . mm/s
If b(t) is the acceleration for a bare head, the area under that curve in the same units is ...
Vb = 1/2(b(2)·2 +1/2(b(2) +b(4))·2 +1/2(b(4) +b(6))·2 +1/2(b(6)·1)
Vb = 1/2(4b(2) +4b(4) +3b(6)) = 1/2(4·120 +4·140 +3·200) = 820 . . . mm/s
Then the difference in speed between the bare head and the helmeted head is ... (0.820 -0.260) m/s = 0.560 m/s
Answer:
2.45 m/s
Explanation:
From the law of conservation of momentum,
The total momentum before the grab = Total momentum after the grab.
mu+m'u' = V(m+m')................... Equation 1
Where m = mass of Erica, m' = mass of Danny, u = initial velocity of Erica, u' = initial velocity of Danny, V = common velocity after garb
Make V the subject of the equation
V = (mu+m'u')/(m+m')............. Equation 2
Given: m = 37 kg, m' = 49 kg, u = 0 m/s(at the maximum height), u' = 4.3 m/s
Substitute into equation 2
V = (37×0+49×4.3)/(37+49)
V = 210.7/86
V = 2.45 m/s
Hence their speed just after the grab = 2.45 m/s
Explanation:
(a) We know that the acceleration of the car is given by :
a = change in speed / time taken
If the speed of the car is constant in a straight line, the acceleration of the car is zero because there is no change in the speed of the car.
(b) For the driver steer a car traveling at constant speed so that the magnitude of the acceleration remains constant, the driver should drive the car in the circular path. This is because, in circular path the speed of an object remains the same while its velocity changes.