The velocity of a Bus is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m.
i. Compute the acceleration.
ii. How much further will the Bus travel before coming to rest, provided the acceleration remain constant?
15t + 1/2 at^2 = 90
Now, a = ∆v/∆t = -8/t
15t - 4t = 90
11t = 90
t = 90/11 s
a = -8/(90/11) = 0.98 m/s^2
so, it will take another 7/0.98 = 7.16 seconds to stop, covering a distance of
7*7.16 - 0.49 * 7.16^2 = 25 meters
can u explain what did u do at the last please? and where did that 0.49 come from?
d=vf*t-1/2 a t^2=7*7.16 - 0.49 * 7.16^2 = 25 meters
i. To compute the acceleration, we can use the following formula:
acceleration (a) = change in velocity (Δv) / time taken (Δt)
The change in velocity is the difference between the final velocity and the initial velocity:
Δv = final velocity - initial velocity = 7 m/s - 15 m/s = -8 m/s
The negative sign indicates that the velocity is decreasing.
The time taken can be calculated using the formula:
Δt = distance / average velocity
Given that the distance is 90 m, we need to find the average velocity. To find the average velocity, we can use the formula:
average velocity = (initial velocity + final velocity) / 2
average velocity = (15 m/s + 7 m/s) / 2 = 22 m/s / 2 = 11 m/s
Now we can calculate the time taken:
Δt = 90 m / 11 m/s ≈ 8.18 seconds
Therefore, the acceleration is:
a = Δv / Δt = -8 m/s / 8.18 s ≈ -0.98 m/s²
ii. To find the distance traveled before the bus comes to rest, we can use the equation of motion:
v² = u² + 2as
where:
v = final velocity (0 m/s, as the bus comes to rest)
u = initial velocity (7 m/s)
a = acceleration (-0.98 m/s², as calculated in part i)
s = distance traveled
Rearranging the equation to solve for s:
s = (v² - u²) / (2a)
Substituting the given values:
s = (0 m/s)² - (7 m/s)² / (2 * -0.98 m/s²)
s = 0 - 49 m²/s² / -1.96 m/s²
s ≈ 24.99 m
Therefore, the bus will travel approximately 24.99 meters further before coming to rest, assuming the acceleration remains constant.