When the brakes of a certain car are applied, a deceleration of 6.0 sec squared is produced. a) How much time will it take to bring the car to a halt from an initial speed of 110km/hr? b) How far will it travel in that time?
If the braking distance is X, and v is the initial velocity, the relationship is
T = V/a
Solve the braking time T. distance X.
For the braking distance X, use
X = Vaverage*T = (V/2)*T = V^2/(2a)
You will have to convert your
To find the time it will take to bring the car to a halt from an initial speed of 110 km/hr, we can use the formula:
v^2 = u^2 + 2as
where:
v = final velocity (0, as the car comes to a halt)
u = initial velocity (110 km/hr, converted to m/s)
a = acceleration/deceleration (-6.0 m/s^2, as negative since it's deceleration)
s = distance traveled
Let's solve part a) first. We will convert the units to SI units (m/s) for consistency:
1 km/hr = 1000 m / 3600 s = 5/18 m/s
So, the initial velocity (u) in m/s is:
u = 110 km/hr * (5/18 m/s)/(1 km/hr) ≈ 30.56 m/s
Now, let's substitute the given values into the formula:
0^2 = (30.56 m/s)^2 + 2 * (-6.0 m/s^2) * s
Simplifying the equation:
0 = 935.9376 m^2/s^2 - 12.0 m/s^2 * s
Rearranging the equation to solve for s:
12.0 m/s^2 * s = 935.9376 m^2/s^2
s = 935.9376 m^2/s^2 / 12.0 m/s^2
s ≈ 77.99 m
Therefore, it will take approximately 77.99 meters to bring the car to a halt from an initial speed of 110 km/hr.
Now, let's move on to part b) to find the distance traveled. Since we already solved for s in part a), the distance traveled is approximately 77.99 meters.