Answer:
I think the answer to your question is true
Answer:
(a) 300 ft
(b) 60 ft/s
Explanation:
distance
where a is acceleration and t is time

Also, d=vt where v is the velocity
d=30t
Therefore
hence t=10 s
Substituting t is either formula
d=30t=30*10=300 ft
Also
v=at hence 
Answer:
2 s, -20 m/s
Explanation:
Given:
y₀ = 20 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t and v
y = y₀ + v₀ t + ½ at²
0 = 20 + 0 + ½ (-9.8) t²
0 = 20 − 4.9 t²
t ≈ 2 s
v² = v₀² + 2a(y − y₀)
v² = 0 + 2(-9.8)(0 − 20)
v ≈ ±20 m/s
Since the rock is falling, v = -20 m/s.
Answer:
a = (v₃₂ - v₂₁) / (t₃₂ -t₂₁)
Explanation:
This is an exercise of average speed, which is defined with the variation of the distance in the unit of time
v = (y₃ - y₂) / (t₃-t₂)
the midpoint of a magnitude is the sum of the magnitude between 2
t_mid = (t₂ + t₃) / 2
the same reasoning is used for the mean acceleration
a = (v_f - v₀) / (t_f - t₀)
in our case
a = (v₃₂ - v₂₁) / (t₃₂ -t₂₁)
Answer:
Work = 165670.4 J = 165.67 KJ
Explanation:
First, we will find the deceleration of the car, using the 3rd equation of motion:

where,
a = deceleration = ?
s = skid distance = 40 m
vf = final speed = 0 m/s
vi = initial speed = 17.2 m/s
Therefore,

the negative sign indicates deceleration here.
Now, we will calculate the braking force applied by the brakes on the car:

the negative sign indicates braking force.
Now, we will calculate the work done using the magnitude of this force:

<u>Work = 165670.4 J = 165.67 KJ</u>