A spring stretches from 10cm to 22cm when a force of 4n is applied, if it obeys Hooke’s law,its total length in cm when a force
1 answer:
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Answer:
Final speed of the train is 7.5 m/s
Explanation:
It is given that,
Uniform acceleration of the train is, a = 1.5 m/s²
It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :
Here, train starts from rest so, u = 0
v = 7.5 m/s
So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.
Answer:
Explanation:
Given
Car speed decreases at a constant rate from 64 mi/h to 30 mi/h
in 3 sec
we know acceleration is given by
negative indicates that it is stopping the car
Distance traveled
s=63.038 m
Answer:
The total distance is 381.5 [m]
Explanation:
In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0
a = acceleration = 2 [m/s²]
t = time = 7 [s]
Vf = 0 + (2*7)
Vf = 14 [m/s]
With this velocity, we can calculate the displacement using the following expression.
where
x = distance traveled [m]
14² = 0 + (2*7*x)
x = 196/(14)
x = 14 [m]
Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.
The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.
We must calculate the final velocity.
Vf = final velocity [m/s]
Vi = initial velocity = 14 [m/s]
a = desacceleration = 4 [m/s²]
t = time = 8 [s]
Vf = 24 + (4*8)
Vf = 56 [m/s]
With this velocity, we can calculate the displacement using the following expression.
where
x = distance traveled [m]
56² = 14² + (2*4*x)
x = 2940/(8)
x = 367.5 [m]
Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.
Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].