The velocity of a Bus is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m.

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.