Till the time car is just adjacent to the bicycle we can say
distance moved by cycle = distance moved by car
Time taken by car to accelerate from rest
![t = \frac{v_f - v_i}{a}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bv_f%20-%20v_i%7D%7Ba%7D)
![t = \frac{49 - 0}{7} = 7 s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B49%20-%200%7D%7B7%7D%20%3D%207%20s)
Time taken by cycle to accelerate
![t = \frac{23 - 0}{15} = 1.53 s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B23%20-%200%7D%7B15%7D%20%3D%201.53%20s)
now the distance moved by cycle in time "t"
![d = \frac{23 + 0}{2}*1.53 + 23(t - 1.53)](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B23%20%2B%200%7D%7B2%7D%2A1.53%20%2B%2023%28t%20-%201.53%29)
distance moved by car in same time
![d = \frac{7t + 0}{2}(t)](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B7t%20%2B%200%7D%7B2%7D%28t%29)
now make them equal
![3.5t^2 = 17.595 - 35.19 + 23t](https://tex.z-dn.net/?f=3.5t%5E2%20%3D%2017.595%20-%2035.19%20%2B%2023t)
![3.5 t^2 - 23t + 17.595 = 0](https://tex.z-dn.net/?f=3.5%20t%5E2%20-%2023t%20%2B%2017.595%20%3D%200)
![t = 5.68 s](https://tex.z-dn.net/?f=t%20%3D%205.68%20s)
so cycle will move ahead of car for t = 5.68 s
Answer:
k = 26.25 N/m
Explanation:
given,
mass of the block= 0.450
distance of the block = + 0.240
acceleration = a_x = -14.0 m/s²
velocity = v_x = + 4 m/s
spring force constant (k) = ?
we know,
x = A cos (ωt - ∅).....(1)
v = - ω A cos (ωt - ∅)....(2)
a = ω²A cos (ωt - ∅).........(3)
![\omega = \sqrt{\dfrac{k}{m}}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cdfrac%7Bk%7D%7Bm%7D%7D)
now from equation (3)
![a_x = \dfrac{k}{m}x](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cdfrac%7Bk%7D%7Bm%7Dx)
![k = \dfrac{m a_x}{x}](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7Bm%20a_x%7D%7Bx%7D)
![k = \dfrac{0.45 \times (-14)}{0.24}](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7B0.45%20%5Ctimes%20%28-14%29%7D%7B0.24%7D)
k = 26.25 N/m
hence, spring force constant is equal to k = 26.25 N/m
Answer:
c it is not accelerating on it's on but gravity pulls it there for velocity increases.
A 59 kg sprinter, starting from rest, runs 47 m in 7.0 s at constant acceleration.?
What is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s?
Instantaneous Power is the force times velocity
P = Fv
Because the acceleration is constant, the force will be constant as well
F = ma
P = mav
for constant acceleration, the velocity at each time is found using
v = at
P = ma(at) = ma²t
find the acceleration using kinematic equation
s = ½at²
a = 2s/t²
a = 2(47) / 7.0²
a = 1.918 m/s²
P(2.0) = 59(1.918²)2.0 = 434.25 W = 0.43 kW
P(4.0) = 59(1.918²)4.0 = 868.51 W = 0.87 kW
P(6.0) = 59(1.918²)6.0 = 1302.76 W = 1.3 kW
I hope this helped.