Answer:
the distance travelled by the traffic, while the car is moving the length of the ramp is 220 m
Explanation:
Given the data in the question;
first we determine acceleration using the kinematic equation below;
v² - u² = 2as
a = v² - u² / 2s
our initial velocity is zero, v is 19 m/s and distance s is 110 m
so we substitute
a = (19² - 0²) / 2×110
a = 361 / 220
a = 1.6409 m/s²
Next, the time t taken by the car to travel along the length of the ramp will be;
t = v - u / a
we substitute
t = (19 - 0) / 1.6409
t = 11.579 sec
so the distance travelled by a body moving with constant speed u is given by following expression:
s = ut + [tex]\frac{1}{2}[/tex] at²
so we substitute 19 m/s for u, 11.579 sec for t and 0 m/s² for a
s = (19 m/s × 11.579 s) + [tex]\frac{1}{2}[/tex] × 0 × (11.579)²
s = (19 m/s × 11.579 s) + 0
s = 220 m
Therefore, the distance travelled by the traffic, while the car is moving the length of the ramp is 220 m
