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3 years ago

A car initially traveling at 17.1 mph comes to rest in 9.7s what was its acceleration in this time?

Physics

2 answers:

slega [8]3 years ago

6 0

Answer: a= -0.79 m/s²

Explanation:solution attached:

First convert 17.1 mi/h to m/s

Conversion factor:

1 mi = 1609.344m

1hr = 3600s

17.1 mi/h x 1609.344 / 1 mi x 1h/ 3600s

= 7.64 m/s

Solve for acceleration:

a= vf - vi / t

= 0 m/s - 7.64 m/s / 9.7 s

= - 0.79 m/s²

The car is decelerating since the result is negative.

ra1l [238]3 years ago

4 0

Answer:

$a=-.78m/s^2$

Explanation:

Δ $v=at$

Δ $v$ is the difference in velocity before and after a given time.
$a$ is the acceleration of the object during this time.
$t$ is time

$(v_f-v_i)=at$ is another way to write this equation.

The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δ $v=(v_f-v_i)$

$v_f-v_i=at\\\frac{at}{t}=\frac{v_f-v_i}{t}\\a=\frac{v_f-v_i}{t}$

I rearranged this equation to solve for $a$ , but this is a step that you don't need to take, it's just good to get in the habit of doing this.
Plug in the given values. Note that our final velocity is $0$ , because the car travels until at rest.

$a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}$

Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.

$a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}$

if you converted correctly, your answer for $v_f$ will be ≅ $7.6m/s$ .
Now divide. Notice that the units for acceleration are $m/s^2$ or meters per second, per second.

$a=\frac{-7.6m/s}{9.7s}\\a=-.78m/s^2$

Our final answer is negative because the car is slowing down. Do not square this answer as the square symbol only applies to the units, not the magnitude.

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SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
For the first collission, only mass 1 is moving before it, so we can write the following equation:

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Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

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From (1) and (2) we get:
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Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

$p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2} = m*v (4)$

Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

$p_{2} = 3*m*v_{2} (5)$

From (4) and (5) we get:
v₂ = v₀/3 (6)

Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

$p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3} = m*v (7)$

Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

$p_{3} = 4*m*v_{3} (8)$

From (7) and (8) we get:
v₃ = v₀/4
This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.

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Answer:

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