Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v = 
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ = 
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Answer:
frequency and amplitude increases
Answer:
the initial velocity of the ball is 104.67 m/s.
Explanation:
Given;
angle of projection, θ = 60⁰
time of flight, T = 18.5 s
let the initial velocity of the ball, = u
The time of flight is given as;
Therefore, the initial velocity of the ball is 104.67 m/s.
Answer:
(A) 7.9 m/s^{2}
(B) 19 m/s
(C) 91 m
Explanation:
initial velocity (U) = 0 mph = 0 m/s
final velocity (V) = 85 mph = 85 x 0.447 = 38 m/s
initial time (ti) = 0 s
final time (t) = 4.8 s
(A) acceleration = 
= 
= 7.9 m/s^{2}
(B) average velocity = 
=
= 19 m/s
(C) distance travelled (S) = ut + 
= (0 x 4.8) + 
= 91 m
Answer:
2.5m/s²
Explanation:
Given parameters:
Mass of car = 200kg
Force on car = 500N
Unknown:
Acceleration of the car = ?
Solution:
According to Newton's second law of motion:
Force = mass x acceleration
Insert the given parameters and find the acceleration;
500 = 200 x acceleration
acceleration = 2.5m/s²
