A car brakes at constant acceleration from a velocity 226m/s to 12.5m/s over a distance of 105m.find;
a)How much time elapses during this time interval.
b)what is the acceleration.
c)if the car continued to be braked at same constant acceleration how much longer would it take for it to stop and what additional distance would it cover?
v=u+at 12.5=226+10t =21.3s
s=ut+0.5at^2 a=20.4
v = Vi + a t
12.5 = 226 + a t
so
a = - 213.5/t
d = Vi t + .5 a t^2
105 = 226 t - .5 * 213/t * t^2
105 = 226 t - 106.5 t
t = .879 seconds
a = -213/.879 = -242 m/s^2
you can do the rest but suspect you have a typo
To find the answers to the given questions, we need to use the equations of motion. Let's break down each question and solve them step by step.
a) How much time elapses during this time interval?
To find the time (t), we can use the equation of motion:
v = u + at
where:
v = final velocity = 12.5 m/s
u = initial velocity = 226 m/s (assuming the car is slowing down)
a = acceleration
Rearranging the equation, we can solve for time (t):
t = (v - u) / a
Substituting the given values:
t = (12.5 - 226) / a
Now, let's move on to the next part to find the acceleration (a).
b) What is the acceleration?
We can use the same equation of motion as in part (a) to find the acceleration (a):
v = u + at
Again, substitute the given values into the equation:
12.5 = 226 + a * t
Solving for acceleration (a):
a = (12.5 - 226) / t
Substitute the value of time (t) from part (a) into this equation to find the acceleration (a).
c) If the car continued to be braked at the same constant acceleration, how much longer would it take for it to stop and what additional distance would it cover?
Since we already calculated the time (t) and acceleration (a) in parts (a) and (b), we can use these values to find the additional time and distance.
To find the additional time, let's assume the final velocity (v2) is 0 (since the car stops):
v2 = u + at
0 = 226 + a * (t + t2) (t2 is the additional time)
To find t2:
t2 = (-226) / a
To find the additional distance:
s = ut + 0.5at^2
Substitute the values of u, a, and t into the equation and calculate s.
Remember to include the negative sign in the formulas for the acceleration (a) and additional time (t2) since the car is decelerating.