A decrease in velocity is referred to as deceleration. If car is moving at 30 m/s and stop in 50 m .The value of deceleration is 11.56 ms−2.
<h3>How to calculate deceleration ?</h3>
While acceleration is motion in which an object's speed varies every second, deceleration is motion that causes an object to slow down.
We are aware that acceleration refers to an object's rate of increase in speed, and deceleration refers to an object's rate of decrease in speed. For instance, when we apply the brakes while driving, we benefit from the vehicle's ability to decelerate and slow down.
The Deceleration Formula is the final velocity minus the initial velocity, with a negative sign in the result because the velocity is decreasing, if starting velocity, final velocity, and time taken are given.
velocity of car = 30 m/s
car need to stop in 50m
Deceleration a = v^2 – u^2 / 2s
= 0^2 - 50^2 / 2*30
= 11.56
Deceleration of the care = 11.56 ms−2
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Answer:
x = 5.79 m
Explanation:
given,
mass of the car = 39000 Kg
spring constant = 5.7 x 10⁵ N/m
acceleration due to gravity = 9.8 m/s²
height of the track = 25 m
length of spring compressed = ?
using conservation of energy
potential energy is converted into spring energy
![m g h = \dfrac{1}{2}kx^2](https://tex.z-dn.net/?f=%20m%20g%20h%20%3D%20%5Cdfrac%7B1%7D%7B2%7Dkx%5E2)
![x =\sqrt{\dfrac{2 m g h}{k}}](https://tex.z-dn.net/?f=x%20%3D%5Csqrt%7B%5Cdfrac%7B2%20m%20g%20h%7D%7Bk%7D%7D)
![x =\sqrt{\dfrac{2\times 39000 \times 9.8 \times 25}{5.7 \times 10^{5}}}](https://tex.z-dn.net/?f=x%20%3D%5Csqrt%7B%5Cdfrac%7B2%5Ctimes%2039000%20%5Ctimes%209.8%20%5Ctimes%2025%7D%7B5.7%20%5Ctimes%2010%5E%7B5%7D%7D%7D)
![x =\sqrt{33.5263}](https://tex.z-dn.net/?f=x%20%3D%5Csqrt%7B33.5263%7D)
x = 5.79 m
the spring is compressed to x = 5.79 m to stop the car.
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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