Consider a car moving along a straight horizontal road with a speed of 72km/h.If the coefficient of static friction between the tyres and the road is 0.5,the shortest distance in which the car can be stopped is(taking g=10m/s^2)
(a)30m (b)40m (c)72m (d)20m
Vi = 72,000 meters/3600 seconds
= 20 m/s
v = Vi + a t
what is a ?
a = F/m
F = -m g (.5)
so
a = -.5 g = -9.81/2 = -4.9 close enough
v = 20 - 4.9 t
when v = 0
0 = 20 - 4.9 t
so t is about 4 seconds
average speed during stop = 20/2 = 10 m/s
so
d = 10 m/s * 4 s = 40 meters
( nearly half a football field )
U - 72km/h
=72×5/18 = 20m/s
V=0
A=-g
=-0.5×10=5m/s^2
V^2=u^2+2as
0=400+2×(-5)×s
0=400-10s
10s=400
S=400/10=40m
To find the shortest distance in which the car can be stopped, we need to determine the deceleration of the car first.
Given:
Initial velocity (u) = 72 km/h = 20 m/s
Coefficient of static friction (μ) = 0.5
Acceleration due to gravity (g) = 10 m/s^2
The maximum static friction force (fs) can be calculated using the formula:
fs = μ * Normal force
The normal force (N) is equal to the weight (mg) of the car, since the car is on a horizontal road.
N = m * g
Now, the deceleration (a) can be calculated using the Newton's second law as:
a = fs / m
To find the stopping distance (s), we can use the following equation of motion:
v^2 = u^2 + 2as
where,
v = final velocity (which is 0, as the car stops)
u = initial velocity
a = deceleration
s = stopping distance
Let's calculate the deceleration first:
N = m * g
N = m * 10
fs = μ * N
fs = 0.5 * m * 10
fs = 5 * m
a = fs / m
a = 5 * m / m
a = 5
Now, let's calculate the stopping distance:
0 = (20)^2 + 2 * 5 * s
0 = 400 + 10 * s
-400 = 10 * s
s = -400 / 10
s = -40
Since distance cannot be negative, the stopping distance cannot be -40m. Therefore, we discard this value.
Therefore, the shortest distance in which the car can be stopped is approximately 40 meters.
Hence, the correct answer is (b) 40m.
To find the shortest distance in which the car can be stopped, we need to calculate the deceleration of the car and then use the equation for distance with constant acceleration.
Given:
Initial speed, u = 72 km/h
Coefficient of static friction, µ = 0.5
Acceleration due to gravity, g = 10 m/s^2
First, we need to convert the initial speed from km/h to m/s:
Since 1 km/h = 1000/3600 m/s
u = (72 * 1000) / 3600 = 20 m/s
The frictional force acting on the car opposes its motion and is equal to the product of the coefficient of friction and the normal force. In this case, the normal force is the weight of the car, which is equal to its mass multiplied by the acceleration due to gravity:
F friction = µ * N = µ * m * g
To calculate the deceleration of the car, we need to find the frictional force. The mass of the car cancels out from both sides of the equation, so we do not need to consider it.
F friction = µ * m * g
Acceleration, a = F friction / m = (µ * m * g) / m = µ * g
Substituting the given values:
a = (0.5 * 10) m/s^2 = 5 m/s^2
Using the equation for distance with constant acceleration:
s = (u^2 - v^2) / (2 * a)
Since the car needs to be stopped, the final speed, v, is 0 m/s:
s = (u^2 - 0^2) / (2 * a) = (20^2) / (2 * 5) = 400 / 10 = 40 m
Therefore, the shortest distance in which the car can be stopped is 40 meters.
Hence, the answer is (b) 40m.