I'm going to cry Need help setting up kinematics problems

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The discussion focuses on solving kinematics problems involving a car's motion with varying acceleration. The first problem involves a car accelerating at 6 m/s² for 5 seconds, maintaining a constant velocity for 10 seconds, and then decelerating at -1 m/s². The second problem examines the horizontal distance a car would land if it fell from different heights. The third problem compares the speeds of two cars traveling at constant velocities, with one decelerating after a specific time. Key equations include those for displacement and velocity under constant acceleration.

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2) Starting from rest, a car accelerates at 6 m/s2 for 5 s, then continues at a constant velocity for 10 s, then decelerates at -1m/s2.
How fast was it going 0.70 km from where it started?



So, a=6m/s2
t= 5 s
ty= 10s
ay= -1m/s2

.70km converted to m is 700m

Where do I go and did I even set it up right? What equations do I use? I know I'm solving for the initial velocity.


3) A car driving straight off a 10m high cliff lands 20 m away. If the cliff had been 30 m high instead of 10 m, how far away would it have landed?


Looking for displacement on this one

So x= 10m
y= 20 m

Do you use pythagorean theorem or try to find v or what? :(



4) Car A and Car B are traveling at constant velocities in the same direction on the highway. Car A passes Car B at 3:07:00 pm. At 3:08:00 pm Car A starts to decelerate at -1 m/s2. At 3:09:00pm Car B passes Car A.
How much faster was Car A driving than Car B when they first met?


...this one I just don't get. I don't even know how to attempt it
 
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You have three periods of time.

0 to 5 seconds, car is accelerating
5 to 15 seconds, car is at constant velocity
15+ seconds, car is decelerating

How fast the car was going when it was 700m away depends on which one of these periods it was in. So that is what you need to work out: in which period does 700m fit?

To determine that, you need to find the distance at 5 seconds and the distance at 15 seconds. Luckily the car doesn't reverse, so you can use those answers to see where 700 fits.

Once you know where 700 fits, you can set about finding the speed.
 
linglingpopo said:
2) Starting from rest, a car accelerates at 6 m/s2 for 5 s, then continues at a constant velocity for 10 s, then decelerates at -1m/s2.
How fast was it going 0.70 km from where it started?
So, a=6m/s2
t= 5 s
ty= 10s
ay= -1m/s2

.70km converted to m is 700m

Where do I go and did I even set it up right? What equations do I use? I know I'm solving for the initial velocity.
Initial Velocity ? The car starts from rest!

Break-up the car's trip into three parts. (Hopefully it travels more than 700m before coming to rest.)

v0=0 m/s , s0 = 0 m , s is displacement, t0 = 0s
Part 1:
From t=t0 to t=t1
a1 = 6 m/s2, t1 = 5 s
What is the velocity, v, at time t during this segment?
What is the velocity, v1, at time t1 at the end of this segment?
How far does the car travel during this segment? - Use average velocity, and s1 = vavgt1 or use s1=v0t1+(1/2)a1(t1)2
Part 2:
From t=t1 to t=t2, where t2-t1 = 10 s .
a2 = __ m/s2, t2 = __ s
What is the velocity, v, at time t during this segment?
What is the velocity, v2, at time t2 at the end of this segment?
How far does the car travel during this segment?
What is the total displacement at time t2 at the end of this segment?
Part 3:
...​
(I'm a slow typist.)
 
Last edited:

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