A car accelerates from rest at 3.00 m/second squared What is the velocity after 5 secs ? What is the displacement after 5 second
1 answer:
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Answer:
a= - 0.79 m/s²
Explanation:
Given that
Speed ,u = 20 mi/h
We know that
1 mi/h= 0.44 m/s
Therefore ,u = 8.94 m/s
Distance ,s= 50 m
Lets take the acceleration of the car = a m/s²
The final speed of the car ,v = 0 m/s
We know that
v²= u² + 2 a s
Now by putting the values
0²= 8.94² + 2 x a x 50
a= - 0.79 m/s²
Therefore the acceleration will be - 0.79 m/s².
To solve this problem we will apply the linear motion kinematic equations. On these equations we will define the speed as the distance traveled in a space of time, and that speed will be in charge of indicating the reaction rate of the individual. In turn, using the ratio of speed, position and acceleration, we will clear the position and determine the distance necessary for braking.
The relation to express the velocity in terms of position for constant acceleration is as follows
Here,
u = Initial velocity
v= Final velocity
a = Acceleration
= Initial position
s = Final position
PART 1) Calculate the displacement within the reaction time
In this case we can calculate the shortest stopping distance
PART 2)
PART 1) Calculate the displacement within the reaction time
In this case we can calculate the shortest stopping distance
While a person without alcohol would cost 517ft to slow down, under alcoholic substances that distance would be 616ft