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

a)compute the acceleration
b) How much farther will the train travel before coming to rest, provided the acceleration remains constant?

To compute the acceleration of the train, you can use the formula:

Acceleration = (Final Velocity - Initial Velocity) / Time

However, we are not given the time it took for the velocity to reduce. Instead, we are given the distance the train traveled while the velocity reduced uniformly.

To find the time, we can use the formula:

Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2

Given:
Initial Velocity (u) = 15 m/s
Final Velocity (v) = 7 m/s
Distance (s) = 90 m

We need to solve for time (t) and acceleration (a) using the above formula. Rearranging the formula, we get:

s = u * t + (1/2) * a * t^2

Substituting the given values:

90 = 15t + (1/2) * a * t^2

Now we can solve this equation to find the values of time and acceleration.

To compute the acceleration, we can use the formula:

acceleration = (final velocity - initial velocity) / time

In this case, the initial velocity (u) is 15 m/s, the final velocity (v) is 7 m/s, and the time (t) is not given. However, we can assume that the time taken for the velocity to reduce uniformly is equal to the distance traveled (s) divided by the average velocity (v_avg).

a) Compute the acceleration:
acceleration = (v - u) / t

Since the train's velocity reduced uniformly, the average velocity can be calculated as:

v_avg = (u + v) / 2

Substituting the values into the equation:

v_avg = (15 m/s + 7 m/s) / 2 = 11 m/s

Now, we can find the time taken (t) using the formula:

t = s / v_avg

Substituting the given distance (s = 90 m) and average velocity (v_avg = 11 m/s):

t = 90 m / 11 m/s ≈ 8.18 seconds

Finally, we can calculate the acceleration:

acceleration = (v - u) / t
acceleration = (7 m/s - 15 m/s) / 8.18 s ≈ -0.98 m/s²

Therefore, the acceleration of the train is approximately -0.98 m/s².

b) To determine how much farther the train will travel before coming to rest, we need to find the distance (s) covered during the deceleration period.

Using the formula:

s = u * t + (1/2) * a * t²

Since the final velocity (v) is 0 m/s when the train comes to rest, the formula can be simplified:

0 = u * t + (1/2) * a * t²

Rearranging the equation:

(1/2) * a * t² = -u * t

Substituting the known values:

(1/2) * (-0.98 m/s²) * t² = -15 m/s * t

Simplifying:

-0.49 m/s² * t² = -15 m/s * t

Dividing both sides by -0.49 m/s²:

t² = (15 m/s * t) / 0.49 m/s²

t ≈ 30.61 seconds

Now, we can find the distance (s) covered during this time using the formula:

s = u * t + (1/2) * a * t²

Substituting the known values:

s = 15 m/s * 30.61 s + (1/2) * (-0.98 m/s²) * (30.61 s)²

s ≈ 230.55 m

Therefore, the train will travel approximately 230.55 meters farther before coming to rest, assuming the acceleration remains constant.