How Do You Calculate Distance Traveled by a Car Under Constant Acceleration?

AI Thread Summary
To calculate the constant acceleration needed for a car to accelerate from zero to 60 mph in 10 seconds, the acceleration is determined to be 2.68 m/s². This acceleration is approximately 27.3% of gravitational acceleration (g = 9.8 m/s²). The distance traveled by the car during this acceleration can be calculated using the equation d = Vi(t) + 0.5(a)(t²), resulting in a distance of 134 meters or 440 feet. The discussion highlights the importance of using the correct equations for solving physics problems. Overall, the thread provides a step-by-step approach to understanding constant acceleration and distance calculations.
ae4jm
Messages
79
Reaction score
0
[SOLVED] Constant Acceleration

Homework Statement


a. What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10s?

b. What fraction of g is this?

c. How far has the car traveled when it reaches 60 mph? Give your answer both in SI units and in feet.


Homework Equations





The Attempt at a Solution


I have gotten answers for parts a and b, but I get really confused on part c.

For a: 1 mph = .447 m/s, so 60(.447)=26.82 m/s. (26.82 m/s)/(10s)=2.68 m/s^2

For b: g=gravity=9.8 m/s^2. (2.68 m/s^2)/(9.8 m/s^2)=.273(100)=27.3%

For c: I'm not sure where to start to find my answer.

I appreciate any help!

I just solved it! I've worked on part C for about 45 minutes to an hour. I just found an equation that worked, imagine that.

For c: I used d=Vi(t)+0.5(a)(t^2) So, d=(0 m/s)(10 s)+0.5(2.68 m/s^2)(10^2)=134 m. Then to convert 134 m to feet = 134(39.37)=5,275.58 in/12=440 ft

Hopefully this will help someone else out that is new to physics, like myself.
 
Last edited:
Physics news on Phys.org
use one of them there equations from your textbook:
<br /> x = \frac{1}{2}a t^2 + v_0 t + x_0<br />
 
This is the exact equation I used! Please see the bottom of my previous post, I just finished editing my initial post. I appreciate your help, all of the great help I've received from this forum!
 
ah. well, that's good.
 
Yeah and hopefully your teacher can help... oh wait...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top