Constant Acceleration Automobile

AI Thread Summary
An automobile accelerates at 2.1 m/s² from a traffic light, while a truck travels at a constant speed of 8.5 m/s, starting 10.2 m ahead. To determine where the automobile overtakes the truck, the equations of motion for both vehicles are set up. After solving the quadratic equation, it is found that the automobile overtakes the truck after approximately 9.16 seconds, covering a distance of 88.1 meters. At that moment, the automobile's speed is 19.2 m/s. The calculations confirm the setup and results are correct.
AnkhUNC
Messages
89
Reaction score
0
[SOLVED] Constant Acceleration

Homework Statement



At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.1 m/s^2. At the same instant a truck is 10.2 m down the road and traveling with a constant speed of 8.5 m/s. (a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automobile be traveling at that instant?

Homework Equations





The Attempt at a Solution



s(t) = 1/2 a * t^2 + v0 * t + s0
a = acceleration
v0 = initial speed
s0 = initial position


Being for the truck
a = 0 m/s^2
v0 = 8.5 m/s
s0 = 10.2 m
and being for the car
a = 2.1 m/s^2
v0 = 0 m/s
s0 = 0 m
the equation will be:
8.5 m/s * t + 10.2 m = (There is a 1/2 here right?) (1/2)2.1 m/s^2 * t^2

Solving this I get 9.156s which is apparently incorrect.
 
Physics news on Phys.org
I believe the equation you set up is correct. Can you show your math?
 
Well the math is kinda already there QQ I just plugged it into quadratic equation.
 
OK I got it!

s(9.16) = 1/2 (2.1) * (9.16)^2 = 88.1 meters

and after 9.16 seconds, it is traveling
v = at
v = 2.1(9.16) = 19.2 m/s
 
Awesome! Good job!
 
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