Answer:
a.5.85 s b i. 15.7 m/s ii. 21.1m/s c. 61.62 m
Step-by-step explanation:
(a) when b will overtake a,
Using s = ut + 1/2at² we express the distance moved by each automobile.
Since both automobiles start at rest, their initial speed, u = 0 m/s, a = acceleration of automobile and t = time of travel or overtaking
So s = 0 × t + 1/2at²
s = 0 + 1/2at²
s = 1/2at²
Let time, t be the time the automobile b start, and its acceleration a = 3.6 m/s²
So, s = 1/2 × 3.6 m/s² × t² = 1.8t²
Since automobile a starts 2 seconds earlier, its time of travel is (t + 2) s. Since its acceleration, a = 2 m/s²,
s = 1/2 × 2 m/s² × (t + 2)²
s = (t + 2)²
Since both distances are equal at overtaking
(t + 2)² = 1.8t²
t² + 4t + 4 = 1.8t²
1.8t²- t² - 4t - 4 = 0
0.8t² - 4t - 4 = 0
dividing through by 0.8, we have
0.8t²/0.8 - 4t/0.8 - 4/0.8 = 0
t² - 5t - 5 = 0
Using the quadratic formula to find t,
t ≅ 5.85 or -0.85
We take the positive answer since t cannot be negative.
So, t = 5.85 s
So, b will overtake a 5.85 s later.
(b) the speed of each automobile at that time,
Using v = u + at for each automobile, where u =initial speed of automobile = 0 m/s (since they both start from rest), a = acceleration of automobile and t = time of travel of automobile
i. Speed of automobile a
For automobile a, its time of travel is (t + 2) s = 5.85 s + 2 s = 7.85 s and its acceleration is 2 m/s²
So, v = u + at
= 0 m/s + 2 m/s² × 7.85 s
= 0 m/s + 15.7 m/s
= 15.7 m/s
ii. Speed of automobile b
For automobile a, its time of travel is t s = 5.85 s and its acceleration is 3.6 m/s²
So, v = u + at
= 0 m/s + 3.6 m/s² × 5.85 s
= 0 m/s + 15.7 m/s
= 21.06 m/s
≅ 21.1 m/s
c) the location where b overtakes a relative to the traffic signal. -truck
The location where b overtakes a relative to the traffic signal is the distance each automobile moves before overtaking. In section (a), we foundthat distance s = 1/2at² = (t + 2)² = 1.8t²
Using s = (t + 2)²
= (5.85 + 2)²
= 7.85²
= 61.6225 m
≅ 61.62 m