Need some help with my homework!

Question 7
a) Acceleration is the rate of change of velocity

Use your definition in (a) to show that v = u + at, where v is the final velocity, u is the initial velocity and a and t are the acceleration and the time interval respectively.

State the conditions that must be satisfied for the equation to be valid.

Question 11
An aircraft has a landing velocity of 50m/s and decelerates uniformly at 10m/s^2 until its velocity is reduced to 10m/s. Calculate
a) The time taken to slow down to this velocity, (answer is 4s but how do I get it?)
b) The distance covered during the deceleration. (answer is 120m but how do I get it?)

Thanks guys! :D

Sure, I would be happy to help you with your homework!

Question 7:
To prove the equation v = u + at, we can use the definition of acceleration: acceleration = change in velocity / change in time. In this case, the initial velocity is u, the final velocity is v, the acceleration is a, and the time interval is t. We want to show that the change in velocity is equal to (v - u), so we have:

acceleration = (v - u) / t

To isolate v, we can multiply both sides of the equation by t and then add u to both sides:

acceleration * t = (v - u)
acceleration * t + u = v

So, v = u + at, as required.

To determine the conditions for this equation to be valid, we need to consider that it applies to an object undergoing constant acceleration. This means that acceleration (a) is constant throughout the entire time interval (t). Additionally, this equation assumes that the initial velocity (u) is known and that the acceleration is acting in the same direction as the change in velocity (v - u).

Question 11:
a) The time taken to slow down to a velocity of 10 m/s can be found using the equation v = u + at, where:
- Initial velocity (u) = 50 m/s
- Final velocity (v) = 10 m/s
- Acceleration (a) = -10 m/s^2 (negative because the velocity is decreasing)

Plug these values into the equation and solve for t:
v = u + at
10 = 50 + (-10)t
10 - 50 = -10t
-40 = -10t
t = -40 / -10
t = 4 seconds

b) The distance covered during the deceleration can be found using the equation of motion s = ut + (1/2)at^2, where:
- Initial velocity (u) = 50 m/s
- Time taken (t) = 4 seconds
- Acceleration (a) = -10 m/s^2 (negative because the velocity is decreasing)

Plug these values into the equation and solve for s:
s = ut + (1/2)at^2
s = 50 * 4 + (1/2) * (-10) * (4^2)
s = 200 - 80
s = 120 meters

So, the time taken to slow down to a velocity of 10 m/s is 4 seconds, and the distance covered during the deceleration is 120 meters.

I hope this helps you with your homework! Let me know if you have any further questions.