Answer:
d = 84 m
Explanation:
As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object
here we know that
initial speed
final speed
time taken by the car to stop
now the distance moved by the car before it stop is given as
now we have
The answer is B) Intentional
Answer:
a) v = 54.7m/s
b) v = (58 - 1.66a) m/s
c) t = 69.9 s
d) v = -58.0 m/s
Explanation:
Given;
The height equation of the arrow;
H = 58t - 0.83t^2
(a) Find the velocity of the arrow after two seconds. m/s;
The velocity of the arrow v can be given as dH/dt, the change in height per unit time.
v = dH/dt = 58 - 2(0.83t) ......1
At t = 2 seconds
v = dH/dt = 58 - 2(0.83×2)
v = 54.7m/s
(b) Find the velocity of the arrow when t = a. m/s
Substituting t = a into equation 1
v = 58 - 2(0.83×a)
v = (58 - 1.66a) m/s
(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s
the time when H = 0
Substituting H = 0, we have;
H = 58t - 0.83t^2 = 0
0.83t^2 = 58t
0.83t = 58
t = 58/0.83
t = 69.9 s
(d) With what velocity will the arrow hit the surface? m/s
from equation 1;
v = dH/dt = 58 - 2(0.83t)
Substituting t = 69.9s
v = 58 - 2(0.83×69.9)
v = -58.0 m/s