Calculate the force needed to bring a 1070–kg car to rest from a speed of 89.0 km/h in a distance of 110 m (a fairly typical distance for a non-panic stop).
Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.34 m. Calculate the force exerted on the car and compare it with the force found in part (a).
FORCE=
RATIO=
Vo = 89 km/h = 89,000m/3600s = 24.72 m/s.
A. V^2 = Vo^2 + 2a*d.
0 = 24.72^2 + 220a.
a = -2.78 m/s^2.
F = M*a = 1070 * (-2.78) = -2972 N. or 2972 N. of opposing force.
B. V^2 = Vo^2 + 2a*d.
0 = 24.72^2 + 4.68a.
a = -130.6 m/s^2.
F = M*a = 1070 * (-130.6) = -139,712 N.
Ratio = 139,712/2972 = 47 times part A.
THANK YOU
HOW WOULD I WRITE THE f & r IN SCIENTIFIC NOTATION
In Scientific notation, the number on the left side of the decimal is always less than 10(a single digit number):
F = 139,712. = 1.39712*10^5.
If I had placed the decimal between 13 and 9, that would not be in scientific notation; because 13 is not a single digit number.
After locating the decimal point, I count the number of decimal places to the right(5). So that is my exponent; it is positive because I counted to the right of the decimal point.
Ratio = 47. = 4.7*10^1.
After locating the decimal, I counted only 1 decimal place to the right. Therefore, the exponent is 1.
For numbers less than 1, the exponent will always be negative:
0.0058 = 5.8*10^-3.
Now look at 0.0058 and count the number of decimal places to the LEFT of your new decimal location(3).
The exponent is negative because we counted to the left of our new decimal point.
I hope this helps.
To calculate the force needed to bring the car to rest, we can use the equation for force:
Force = mass * acceleration
First, let's convert the speed from km/h to m/s:
89.0 km/h = (89.0 * 1000) m / (3600) s ≈ 24.7 m/s
Part (a):
In this case, the car is brought to rest in a distance of 110 m. To calculate the force, we need to find the acceleration using the equations of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, since the car comes to rest)
u = initial velocity (24.7 m/s)
a = acceleration
s = distance (110 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
= (0^2 - 24.7^2) / (2 * 110)
a ≈ -2.99 m/s^2 (the negative sign indicates deceleration)
Now, we can calculate the force:
Force = mass * acceleration
= 1070 kg * (-2.99 m/s^2)
Force ≈ -3,203 N
Part (b):
In this case, the car is brought to rest in a distance of 2.34 m. Similarly, we need to find the acceleration using the equations of motion:
a = (v^2 - u^2) / (2s)
= (0^2 - 24.7^2) / (2 * 2.34)
a ≈ -671 m/s^2 (the negative sign indicates deceleration)
Now, we can calculate the force:
Force = mass * acceleration
= 1070 kg * (-671 m/s^2)
Force ≈ -717,170 N
The force exerted on the car when hitting the concrete abutment at full speed is much larger (about 223 times) than the force required to bring the car to rest in a typical non-panic stop.