You're driving at 39km/h , when the traffic light 25m away turns yellow.
I need to find the constant acceleration required to stop at the light, the stopping time, and weather the acceleration is reasonable.
I know how to find the average and instantaneous acceleration, but how to obtain the constant, I have no clue.
39,000 m/3600 s = 10.83 m/s
average speed during braking = 10.83/2 = 5.42
25 = 5.42 t
t = 25/5.42 = 4.62 seconds to stop
v = Vi + a t
0 = 10.83 + a (4.62)
a = -2.34 m/s^2
1/4 g is not an unreasonable deacceleration
39 km/hr = 10.83 m/s
s = Vi*t + 1/2 at^2
25 = 10.83t + .5 at^2
v = Vi + at
10.83 + at = 0
so,
t = 2.35s
a = -4.46 m/s^2
Go with Damon - I switched the a and the t.
To find the constant acceleration required to stop at the traffic light, you can use the kinematic equation for distance:
d = (v_i^2 - v_f^2) / (2a),
where:
d is the distance,
v_i is the initial velocity,
v_f is the final velocity, and
a is the constant acceleration.
Given:
v_i = 39 km/h,
v_f = 0 (final velocity is 0 since you want to stop),
d = 25 m (convert to km: 25 m ÷ 1000 = 0.025 km),
First, convert the initial velocity to m/s by using the formula:
v_i_mps = v_i km/h × (1000 m/1 km) × (1 h/3600 s).
Then, substitute the values into the formula:
0.025 km = (v_i_mps^2 - 0^2) / (2a).
To isolate 'a', you can solve for it algebraically:
2a × 0.025 km = v_i_mps^2.
To get 'a', divide both sides by (2 × 0.025 km):
a = v_i_mps^2 / (2 × 0.025 km).
Once you've obtained the value of 'a', you can calculate the stopping time using the equation:
t = (v_f - v_i) / a.
Substitute the values for final velocity (v_f), initial velocity (v_i), and the value of 'a' you just calculated into the equation to find the stopping time.
To determine if the acceleration is reasonable, compare it to typical values for vehicle acceleration. For example, the average car's maximum acceleration is usually around 3-4 m/s^2. If the calculated acceleration is within this range or similar to the expected values, it can be considered reasonable.