What is the acceleration of the car

  • Thread starter Thread starter colin_delecia
  • Start date Start date
  • Tags Tags
    Acceleration Car
AI Thread Summary
The problem involves a car decelerating from 25 m/s to rest over a distance of 75 meters. To find the acceleration, the equation v² = u² + 2as is used, where v is the final velocity (0 m/s), u is the initial velocity (25 m/s), and s is the distance (75 m). Solving for acceleration gives a result of -0.16 m/s², indicating that the car is decelerating. The negative sign confirms that the acceleration is in the opposite direction of the initial velocity. This approach effectively eliminates the need for time in the calculation.
colin_delecia
Messages
4
Reaction score
0

Homework Statement



A car slows down from 25 to rest in a distance of 75 .
What is the acceleration of the car.

I've been fiddling with this problem pretty much all night but can't seem to get to a correct answer, help would be apprechiated.

Homework Equations



x = x0 + v0t + 1/2at^2
d = 1/2(v0 +v)t

The Attempt at a Solution



x = x0 + v0t + 1/2at^2

75m = (25m/s)(60.5s) + 1/2a(60.5s)^2

75m = 1512.5m + 1/2a(60.5s)(60.5s)
75m = 1512.5m + 1/2a(3660.25s^2)

75m - 1512.5m = 1512.4m - 1512.4m 1/2a(3660.25s^2)
-1437.5m = .5a(3660.25s^2)
------ ----------------

.5 .5


-2875 = a(3660.25s^2)
------ -------------
3660.25 3660.25s^2


-.79m/s2

AND

-25m/s
------
60.5s = -.42m/s2
 
Last edited:
Physics news on Phys.org


em i don't understand how you got your time? anyway

notice this question doesn't give you time.

so you use the equation of motion that doesn't involve time, which is

v2 = u2 + 2as

now , v2 = 0 , because final car velocity = 0
u2 = 25 as stated in question
s = 75

so 0 = 25 + 2a(75)
a = -25/(2x75) = -0.16

so the negative sign tells you the acceleration is opposite to the direction of the +25 m/s velocity , i.e, the car is decelerating .
 
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