A car is moving at 80km/h is subjected to a uniform retardation and brought to rest in 70seconds.find retardation and distance travelled by the car before it comes to rest.
Vo = 80,000m/3600s = 22.22 m/s.
V= Vo + a*t = 0.
22.22 + a*70 = 0,
a = -0.317 m/s^2.
V^2 = Vo^2 + 2a*d = 0.
(22.22)^2 - 0.63779 4d = 0,
d = 779 m.
Correct answer of distance is 2332.5m
To find the retardation and distance traveled by the car, we will use the equation of motion:
v = u + at
Where:
v = final velocity (0 m/s, as the car comes to rest)
u = initial velocity (80 km/h)
a = acceleration (retardation)
t = time taken (70 seconds converted to hours)
First, let's convert the initial velocity from km/h to m/s. To do this, we need to remember that 1 km/h is equal to 1000 m/3600 s.
So, 80 km/h = (80 * 1000) / 3600 = 80000 / 3600 ≈ 22.22 m/s
Now, let's plug in the values into the equation of motion:
0 = 22.22 + a * 70
Simplifying the equation:
-22.22 = 70a
Now, divide both sides of the equation by 70:
a = -22.22 / 70 ≈ -0.317 m/s²
The negative sign indicates that the acceleration (retardation) is in the opposite direction to the initial velocity.
Now, let's find the distance traveled by the car before it comes to rest. We can use the formula:
s = ut + (1/2)at²
Where:
s = distance traveled
u = initial velocity (22.22 m/s)
t = time taken (70s)
a = acceleration ( -0.317 m/s²)
Plugging in the values:
s = (22.22 * 70) + (1/2)(-0.317)(70²)
Simplifying the equation:
s = 1555.4 + (-7.0215) ≈ 1548.3785 m
Therefore, the retardation of the car is approximately -0.317 m/s² and the distance traveled by the car before it comes to rest is approximately 1548.3785 meters.
Iam not satisfy
vf=vi+a*t
but a= force/mass
vf=vi+force*time/mass
So, you need the mass of the car to determine force of slowing down.