The Physics Problem Solver The motion of a vehicle

The Physics Problem Solver The motion of a vehicle on the highway Helmi Abdullah Program Studi Pendidikan Fisika Universitas Negeri Makassar-INDONESIA Email. [email protected]

Problem-1 The car first stays at the toll gate, then moves up to the acceleration is 2m/ s2. Then move constant for 150s, after that, the driver slows down the speed is 1m/s2 until it stops. If the total distance traveled by the car is 7200m specify how long the car is moving?

Solution Problem-1 by Sketch of Knowledge Strategy This question relates to the average speed. The mechanism for solving this problem is by using a sketch of knowledge strategy. The first stage is the statement of the problem is translated into a sketch. The sketch is:

v

v

a1

a2

S1

S3

150s 7200m Fig.1.1

Seen in figure-1.1, that there are three stages of motion of the car are: (1) a = 2m / ss, (2) move constant for 150s, and (3) a = -1m / s2. For accelerated car movement the equation is: S1 =

1 2 (t1 )2 = t12 2

… (1.1)

For car movement with constant speed for 150s the equation is: S2 = 150v

… (1.2)

Helmi Abdullah, Program Studi Pendidikan Fisika Universitas Negeri Makassar, 2018

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As for the slowed car movement, the equation is S3 =

1 1 t2 2

2

=

1 2 t 2 2

… (1.3)

And also known that the total distance of the car is 7200m, then the relationship S1, S2, and S3 formulated as follows. 1 7200 = S1 + S2 + S3 = t12 + 150v + t 22 2 or 2t12 + 300v + t 22 = 14400

… (1.4)

In equation (1.4) there is still a variable to be determined that is the values of v and t2. If you see the sketch above, then the car speed at the time t 1 equal to the speed of the car when t2 so it can be formulated: v = 2 (t1 ) = 1 t 2 or t 2 = 2t1

(1.5)

If equation (1.5) is substituted to equation (1.4) then 3t12 + 300t1 − 7200 = 0 and t1 = 20s and t 2 = 2 t1 = 40s Thus the total distance traveled the car is t = 20 + 150 + 40 = 210s

Solution Problem-1 by Graphical strategy The above problems can also be solved by graph strategy. The graphs used are graphs v and t. Where the area of the motion equation curve bounded by the t axis is to express the distance traveled by the car. For the above matter, the graphical form of the motion equation is:


2

v(m/s)

v=2t1=t2

2m/s2

1m/s2

t(s)

0 150s

t1

t2

Fig.1.2

Based on the graph in figure-1.2, that shaded part of the curve is the total distance traveled by the car that is 7200m. Then the shaded curve area is formulated:: 7200 =

1 t + 150 + t 2 2t1 2 1

… (1.6)

and from the graph of figure 2 above the speed of the car shown on the v-axis is v=2t1 = t2, then the equation (1.6) becomes: 3t12 + 300t1 − 7200 = 0 or t12 + 100t1 − 2400 = 0

… (1.7)

Equation (1.7) is a quadratic equation that has two solutions, and the correct solution is: t = 210s

Problem-2 A car is moving at a constant speed of 10m/s on the highway. When t = 0, the car accelerates its speed by accelerating 0.5m/s2 during the 40s, then slowing down because at a distance of 200m there is a car crash. If the car stops right at the point of the collision then determine the average speed of the car. Solution Problem-2 by Sketch of Knowledge Strategy


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The sketch of problem-2 is shown in figure-2.1 as follows

10m/s

Mobil berhent i

v a

0,5m/s2 S

200m

t

40s Fig.2.1

From figure-2.1, there are two stages of the movement of the car. The first stage is accelerated cars for 40s. The second stage is the car slowed to a stop during t. The total distance traveled during the move to stop is (S + 200) m, with the total time being (40 + t). Thus the average speed of the car during the move is: v=

S + 200 40 + t

… (2.1)

There are two variables that have not been known in equation (2.1) that is S and t. To find the value of S then look again picture-2.1 above. The S variable is at an accelerated car stage. Then the distance S is S = 10 40 +

1 0,5 402 = 800m 2

… (2.2)

Then for the value of t obtained from the motion of the car slowed with the initial velocity v to stop. The equation for this slowed motion is 0 = v − at or t=

v a

… (2.3)

So there are two variables that must be determined that is v and a. For v can be determined from the movement of the car in the previous stage, because v is nothing but the final speed for the car when accelerated, so obtained: m v = 10 + 0,5 40 = 30 … (2.4) s As for the slowed motion of the car are: v 2 = 2a 200 = 400a

… (2.5)


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by substituting the value of v from equation (2.4) to equation (2.5) it is obtained: a=

302 = 2.25 m/s 2 400

thus the equation (2.3) becomes: t=

v 30 = ≈ 13.33s a 2,25

then the average speed of the car is: v=

1000 = 18.75 m/s 53,33

Solution Problem-2 by Graphical strategy The shape of the graph curve for problem-2 is

v(m/s)

v= 0,5(40)+10

30

0,5m/s2

10

200m t(s) 0 t

40s Fig.2.2

Based on figure-2.2. above, then the average speed of the car is obtained by calculating the area of shaded area then divided by the total time, then: 1 30 + 10 40 + 200 1000 v=2 = 40 + t 40 + t

… (2.6)

and "t" can be obtained from the equation as follows: t=

200 40 = s 1 3 (30) 2


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then the average speed of the car is: v=

1000 3000 = = 18,75m/s 40 160 40 + 3

Problem-3 At first the car was stationary at the entrance gate of the highway. Then the car moves with acceleration 10m/s2 up to speed reaches 30m/s. After that, the car is moving at a constant speed. After some time moving at a constant speed, the car moves with a slowdown of 5m/s2 until it stops right at the exit gate. If the average speed of the car is 20m/s, then determine how long the car is moving at a constant speed. Solution Problem-3 by Sketch of Knowledge Strategy Based on the problem-3 statement can be determined that there are three stages of the movement of the car namely: (1) the car is moving accelerated, (2) the car is constantly moving, and (3) the car moves slowed to stop. The sketch of the problem3 is as follows..

30m/s

30m/s

10m/s2 5m/s S1

S2

t1

berhenti

2

S3 t3

t2

Fig.3.1

From figure-3.1 it can be determined that for t=t1 the motion equation of the car is: 302 = 2 10 (S1 ) and S1 =

900 = 45m 20

and t1 =

2(45) = 3s 10

For t = t2 the car is moving constant, then the equation is: Helmi Abdullah, Program Studi Pendidikan Fisika Universitas Negeri Makassar, 2018

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S2 = (30)t 2 For t = t3, the car moves slowed, then the equation is: S3 =

v2 302 = = 90m 2a 2(5)

and t3 =

2(90) = 6s 5

Thus the average speed of the car is: 20 =

S1 + S2 + S3 135 + 30t 2 = t1 + t 2 + t 3 9 + t2

and t 2 = 4,5s So the time it takes the car to move at a constant speed is 4.5 s. Solution Problem-3 by Graphical strategy The graph curve of problem-3 is illustrated as follows.

v(m/s)

30=10t1=5t3

10m/s2

5m/s2

0 t1

t2

t(s)

t3 Fig.3.2

From figure-3.2, the curve area below the graph is the total distance traveled by the car, so the average speed of the car during the move to stop is: 1 t1 + t 2 + t 3 + t 2 (30)] 20 = 2 t1 + t 2 + t 3 Helmi Abdullah, Program Studi Pendidikan Fisika Universitas Negeri Makassar, 2018

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and t2 =

t1 + t 3 2

… (3.1)

On the -v axis is obtained: 30 = 10t1 = 5t 3

… (3.2)

from equation (3.2) we get the value of t1 = 3s and t3 = 6s, so equation (3.1) is the time it takes the constant moving car is: t 2 = 4,5s


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