Bredan is driving through town at 25m/s and begins to accelerate at a constant rate of -1.0m/s/s. Eventually Brendan comes to a complete stop.

A) Represent Brendan's accelerated motion by sketching a velocity-time graph.
B) Use the velocity-time graph to determine his total displacement.
C) Use kinematic a equations to calculate the displacement which Brendan travels while accelerating.

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The distance traveled can be found by calculating the area between the line on the graph and the axes.

Area = 0.5*bh
Area = 0.5*(25.0 s)*(25.0 m/s)
Area = 313 m

A) To sketch Brendan's accelerated motion on a velocity-time graph, we start with a horizontal line at 25 m/s to represent his initial velocity. Since he is decelerating at a constant rate of -1.0 m/s², the line must slope downwards. As time progresses, the velocity decreases until it reaches zero, which indicates that Brendan has come to a complete stop. This results in a straight line with a negative slope from the initial velocity of 25 m/s until it intersects the time axis at zero velocity.

B) To determine Brendan's total displacement using the velocity-time graph, we need to find the area under the graph. Since Brendan's motion starts with a positive velocity (25 m/s) and ends with zero velocity, we can calculate the displacement by finding the area of the triangle formed by the velocity-time graph. The formula to find the area of a triangle is A = (base * height) / 2. In this case, the base is the time taken to come to a stop, and the height is the initial velocity of 25 m/s.

C) Now, let's calculate the displacement using kinematic equations. We can use the equation:

v² = u² + 2as

where
v = final velocity (0 m/s - since he comes to a stop)
u = initial velocity (25 m/s)
a = acceleration (-1.0 m/s²)
s = displacement

Rearranging the equation to solve for s, we have:

s = (v² - u²) / (2a)

Substituting the given values, we get:

s = (0² - 25²) / (2 * (-1.0))

Calculating the expression:

s = (0 - 625) / (-2)

s = 625 / 2

s = 312.5 m

Therefore, Brendan's displacement while accelerating is 312.5 meters.