You will be surprised by this car's speed!!!!

Classical Mechanics Level 2

A car moves with constant velocity 2 m . s − 1 2 m.s^{-1} for 5 second. After 5 second, the car accelerates as much as 10 m . s − 2 10 m.s^{-2} . After a while, the car then start slowing with negative acceleration as much as 10 m . s − 2 10 m.s^{-2} until it stop. If the car moves for 15 second, find out the total distances of the car's move (in m m )!

290.1 250.4 299.1 269.9 258.5

Andronikus Lumembang by Andronikus Lumembang , 25, Indonesia

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1 solution

While slowing down, it reaches a velocity of 2m/sec.
This velocity is destroyed by deceleration of 10m/sec/sec.
The time taken is =2/10 sec= 0.2 sec. Distance traveled =mean velocity * time =
2/2 * 0.2= 0.2 m. ...............(1) ........ .. .

First 5 sec were with velocity 2m/sec and distance traveled = 2*5=10 m. ...(2) . . .

Thus the time taken for accelerating and decelerating = 15-5-.2= 9.8 sec.
For 9.8 sec the mean velocity = 2+(10 9.8/2 + 0)/2=26.5 m/sec.
Distance traveled with 26.5 m/sec = 26.5 9.8= 259.7 m

Total distance traveled= 10 +259.7 +.2 = 269.9 m ~~~\boxed{ 269.9 m }

The distance travelled is broken in to three parts.

(1) Where 2m/sec is acting. R E D a r e a a r e c t a n g l e . \color{#D61F06}{RED ~area}~a~rectangle.

(2)Where there is equal acceleration and deceleration up till the velocity is 2m/sec. . . B L U E a r e a 2 t r i a n g l e s . \color{#624F41}{ \text{up till the velocity is 2m/sec.}}.. \color{#3D99F6}{ BLUE~~area~2~triangles.}

(3)Where the deceleration -10m/sec/sec reduces the 2m/sec velovity to 0 m/sec in time t sec. G R E E N a r e a a t r i a n g l e . \color{#20A900}{GREEN~~area~a~triangle.}

(3)t= velovity/ deceleration=2/10 =0.2sec.~~ The distance travelled in
0.2sec = 1 2 ∗ 0.2 ∗ 2 = 0.2 m =\dfrac{1}{2}*0.2*2=\color{#20A900}{0.2m}
(1)2m/sec acts for 15 – 0.2 = 14.8 sec and travaelles 2 ∗ 14.8 = 29.6 m . 2*14.8=\color{#D61F06}{29.6m.}
(2)Time for equal 10m/sec/sec acceleration and deceleration acts= total time 2m/sec acts - time 2m/sec acts alone = 14.8 – 5= 9.8 = 4.9 + 4.9 sec.
This changes the velocity by 10 * 4.9 =49m/sec. Therefore the distance travelled, the blue area = 2 ∗ 1 2 ∗ 4.9 ∗ 49 = 240.1 m 2*\dfrac{1}{2}*4.9*49= \color{#3D99F6}{240.1 m}

Total distance travelled =Three colored area=0.2+29.6+240.1= 269.9 \\\boxed {\huge{ 269.9} }

Ujjwal Rane has several graphic solutions in Brilliant and You tube.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Yup, use the same method :D

Andronikus Lumembang - 6 years, 4 months ago

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