A race car can be slowed with a constant acceleration of -11 m/s^2
a. If the car is going 55 m/s, how many meters will it travel before it stops?
b. How many meters will it take to stop a car going twice as fast?
According to one of the teachers this is what they stated
A. stopping distance = (stopping time) x (average speed)
= (V/a)*(V/2) = V^2/(2a)
B. If the speed doubles, with deceleration rate "a" staying the same, the stopping distance is four times farther.
Ok, to find the distance for the first one i understand. I'm pretty sure this is how you do it
d = Vit
d = 55 x 5
d = 275 m
But for b, Would i have to double the speed for this which is 55 x 2 = 110 and would i also have to increase the stoppping distance which is 275 m by multiplying by 4
This question has been answered twice already.
Yes, the stopping distance quadruples if you double the speed.
What is there about
Stopping distance = V^2/(2a)
that you don't understand?
Ok, i just understand this a little. I know it has been answered twice by you and i thank you for that. Can you tell me if this right below?
a. stoppping distance = v^2 / 2a
sd = 55^2 / 2(- 11)
sd = 3025 / - 22
sd = -137.5 m
b. Therefore, if the stopping distance quadruples then i shouldn't put that in the equation of:
sd = v^2 / 2a
sd = 110^2 / 2 (- 11)
sd = 12100 / -22
sd = -550 m
Is this right? Thanks i appreciate this a lot and happy new yr.
Well, let's see. To calculate the stopping distance for the car going twice as fast, we can use the formula you mentioned: stopping distance = V^2/(2a).
For part a, we have V = 55 m/s and a = -11 m/s^2. Plugging in these values, we get:
stopping distance = (55^2)/(2*(-11))
= 3025/-22
≈ -137.95 m
Hmm, that's a negative distance. I guess the car started going backwards when it was trying to slow down. Let's just take the absolute value of this distance and say the car traveled approximately 138 meters before stopping.
Now, for part b, if we double the speed (to 110 m/s) but keep the same acceleration, the stopping distance quadruples. So, we take the original stopping distance (275 m) and multiply it by 4:
new stopping distance = 275 * 4
= 1100 m
Therefore, it would take approximately 1100 meters for the car, going twice as fast, to come to a stop. Just remember, no matter how fast the car is going, don't be a clown and be sure to brake safely!
To find the stopping distance for the first question, you correctly used the equation d = Vit, where d is the distance, Vi is the initial velocity, and t is the time. In this case, the initial velocity is 55 m/s, and the acceleration is -11 m/s^2. To find the time it takes for the car to stop, you can use the equation a = (Vf - Vi) / t, where a is the acceleration, Vf is the final velocity (which is 0 m/s when the car comes to a stop), and Vi is the initial velocity. Rearranging the equation, you get t = (Vf - Vi) / a. Plugging in the values, you get t = (0 - 55) / -11 = 5 seconds.
Now, you can substitute the values into the equation d = Vit to find the distance traveled. d = 55 m/s * 5 s = 275 m, which matches your calculation.
For the second question, if the car is going twice as fast, the initial velocity will be 2 * 55 m/s = 110 m/s. However, the deceleration rate "a" remains the same at -11 m/s^2.
According to the teacher's statement, if the speed doubles, the stopping distance is four times farther. This means that you don't need to multiply the stopping distance by 4 since that is already accounted for in the equation.
Using the equation d = V^2 / (2a), substitute the values to find the stopping distance. d = (110 m/s)^2 / (2 * -11 m/s^2) = 6050 m / -22 m/s^2 = -275 m.
The negative sign in the calculation indicates that the stopping distance is in the opposite direction of the car's initial motion. However, since distance cannot be negative, you take the absolute value of the result, which gives you 275 m.
Therefore, the stopping distance for a car going twice as fast is the same as the stopping distance in the first question, which is 275 m.