On turning a corner a car drIver driving at 36 km per hour find a child on the road 55 metre ahead.he immediately applies brakes so as stop within 5 metre. Calculate the retardation produced and the time taken by the car to stop.
To calculate the retardation produced, we need to use the following formula:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car comes to a stop)
u = initial velocity (36 km/h = 10 m/s)
a = acceleration (retardation)
s = displacement (55 m – 5 m = 50 m)
Rearranging the formula, we have:
0 = (10 m/s)^2 + 2a(50 m)
Simplifying further:
0 = 100 m^2/s^2 + 100a m
Subtracting 100a m from both sides:
-100a m = 100 m^2/s^2
Dividing by -100 m:
a = -1 m/s^2
Therefore, the retardation produced by the car is 1 m/s^2.
To calculate the time taken by the car to stop, we can use the formula:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (10 m/s)
a = acceleration (retardation, -1 m/s^2)
t = time
Rearranging the formula, we have:
0 = 10 m/s - 1 m/s^2 * t
Simplifying further:
1 m/s^2 * t = 10 m/s
Dividing by 1 m/s^2:
t = 10 s
Therefore, the car takes 10 seconds to come to a stop.
To find the retardation produced and the time taken by the car to stop, we can use the equations of motion.
First, let's convert the speed of the car from km per hour to meters per second (m/s):
Speed of the car = 36 km/h = (36 * 1000) / (60 * 60) m/s = 10 m/s (approximately)
Given:
Initial velocity (u) = 10 m/s (since the car was already moving at this speed)
Final velocity (v) = 0 m/s (since the car needs to stop)
Distance traveled (s) = 55 m (the distance between the car and the child)
Deceleration or retardation (a) = ?
Using the equation of motion:
v^2 = u^2 + 2as
Substituting the given values:
0^2 = (10)^2 + 2 * a * 55
Rearranging the equation:
0 = 100 + 110a
110a = -100
Dividing both sides by 110:
a = -100/110
a ≈ -0.91 m/s^2
The retardation produced is approximately -0.91 m/s^2.
Next, let's find the time taken by the car to stop.
Using the equation of motion:
v = u + at
Substituting the given values:
0 = 10 + (-0.91) * t
Rearranging the equation:
0 = 10 - 0.91t
0.91t = 10
Dividing both sides by 0.91:
t ≈ 10/0.91
t ≈ 10.99 seconds
The time taken by the car to stop is approximately 10.99 seconds.
vf^2=vi^2+2ad
solve for a. Now I am not certain if by "retardation" you mean a, or the force braking, if force, then force= a/masscar
time:
timetostop= distance/avgveloctiy
= distance/(vf/2)