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3 years ago

6

An object initially at rest accelerates at 5 meters per second2 until it attains a speed of 30 meters per second. What distance

does the object traverse while accelerating?

2 answers:

Misha Larkins [42]3 years ago

6 0

Answer:

Distance = speed * time

the formula S =ut +1/2at^2 or S =( v +u)/2*t

where u is initial velocity and v is final velocity

where a is acceleration due to gravity and where t is time

Explanation:

Tcecarenko [31]3 years ago

4 0

Answer: 90m

Explanation:

Use Equation for distance:

S=a*t²/2

use eqation for acceleraton a=(Vf-Vs)/t

Vs- starting speed

Vf - final speed

a-accelaration

---------------------------------

a=5m/s²

Vs=0m/s

Vf=30m/s

use

a=(Vf-Vs)/t

to find time

t=(Vf-Vs)/a

t=30m/s/5m/s²

t=6s

Now calculate distance that object travel using:

S=(a*t²)72

s=(5m/s²*(6s)²)/2

S=90m

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a)

Generally, the equation for the force of equilibrium is mathematically given as

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Therefore

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By force equilibrium

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Therefore

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c)

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By force equilibrium

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Harlamova29_29 [7]

Answer:

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Explanation:

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$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

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