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bogdanovich [222]
3 years ago
5

Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and

then comes to stop in 7s.
A) calculate stan’s accelerator for each segment of the motion.

b) find the result displacement after 26s.

c) calculate stan’s average velocity
Physics
1 answer:
kow [346]3 years ago
7 0

A) Acceleration: a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

A)

Acceleration is defined as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first  segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

a_1 = \frac{11-8}{4}=0.75 m/s^2 (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: a_2 = 0

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

a=\frac{0-11}{7}=-1.57 m/s^2

And the negative sign means the acceleration is south, opposite to the direction of motion.

B)

In a uniformly accelerated motion, the displacement can be calculated as:

s=ut+\frac{1}{2}at^2

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

u = 0\\a = 0.75 m/s^2\\t=4 s

So the displacement is

s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m

- For the second segment, we have

u = 11 m/s\\a = 0\\t=15 s

So the displacement is

s_2 = (11)(15)+0=165 m

- For the third segment, we have

u = 11\\a = -1.57 m/s^2\\t=7 s

So the displacement is

s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

C)

The average velocity is given by:

v=\frac{d}{t}

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

v=\frac{209.5}{26}=8.06 m/s (north)

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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Read 2 more answers
(a) Calculate the height of a cliff if it takes 2. 35 s for a rock to hit the ground when it is thrown straight up from the clif
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(a) The height of the cliff will be 8.26 meters.

(b) The time would it take to reach the ground will be 0.717 sec.

<h3>What is velocity?</h3><h3 />

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component.

(a) The height of the cliff will be 8.26 meters.

According to Newton's second equation of motion

\rm H =ut-\frac{1}{2} gt^2 \\\\ \rm H =8\times 2.35-\frac{1}{2} 9.81 (2.35)^2\\\\\rm H =8.16 \; m

Hence The height of the cliff will be 8.26 meters.

(b)The time would it take to reach the ground will be 0.717 sec.

We must have the final velocity to find the time so;

\rm v^2=u^2+2gh\\\\ \rm v^2=8^2+2\times 9.81 \times 8.6 \\\\ \rm v= \sqrt{8^2+2\times 9.81 \times 8.6}\\\\\rm v=15.03 \;m/sec

According to Newton's third equation of motion ;

\rm v=u-gt \\\\ \rm t=\frac{v-u}{g} \\\\ \rm t=\frac{15.03-8}{9.81} \\\\ \rm t=0.717 sec.

Hence the time would it take to reach the ground will be 0.717 sec.

To learn more about the velocity refer to the link ;

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