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3 years ago

A ball rolled along a horizontal surface comes to rest in a distance of 72m in 6s. Its initial velocity

Physics

1 answer:

Darina [25.2K]3 years ago

3 0

Answer:

1. Initial velocity = 24 m/s

2. Deceleration = –4 m/s²

Explanation:

From the question given above, the following data were obtained:

Distance travelled (s) = 72 m

Time (t) = 6 s

Final velocity (v) = 0 m/s

1. Determination of the initial velocity.

Distance travelled (s) = 72 m

Time (t) = 6 s

Final velocity (v) = 0 m/s

Initial velocity (u) =?

s = (u + v)t / 2

72 = (u + 0) × 6 / 2

72 = u × 3

Divide both side by 3

u = 72 / 3

u = 24 m/s

2. Determination of the deceleration.

Time (t) = 6 s

Final velocity (v) = 0 m/s

Initial velocity (u) = 24 m/s

Deceleration (a) =?

v = u + at

0 = 24 + (a × 6)

0 = 24 + 6a

Collect like terms

0 – 24 = 6a

–24 = 6a

Divide both side by 6

a = –24 / 6

a = –4 m/s²

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3 years ago

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Explanation:

Given

Length traveled by the ball $s=40\ m$

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Initial velocity $u=0\ m/s$

Using the second equation of motion

$\Rightarrow s=ut+\frac{1}{2}at^2\\\Rightarrow 40=0+\frac{1}{2}a(3)^2\\\Rightarrow a=\frac{80}{9}\ m/s^2\\$

Now using

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A girl and her bicycle have a total mads of 42 kg. At the top of the hill her speed is 4 m/s. The hill is 14.3m high and 112m lo

frosja888 [35]

Answer:

13.78 m/s

Explanation:

Given that:

The mass of the girl & her bicycle = 42 kg

The speed at the top of the hilll= 4 m/s

The height of the hill = 14.3 m

The length of the hill (distance) = 112 m

The frictional force = 20 N

To find the speed at the bottom of the hill; we need to carry out the following processes.

The workdone by gravity = mass × acceleration due to gravity × Δh

= 42 × 9.8 × ( 14.3 - 0 )

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The workdone by the friction = Force × distance = - 20 × 112 (since she is riding down the hill)

The workdone by the friction = -2240 Joules

The initial Kinetic friction = 1/2 mv²

= 1/2 × 42 × 4²

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The final kinetic energy = Initial Kinetic energy + total work

The final kinetic energy = (336 + 5885.88 - 2240) Joules

The final kinetic energy = 3981.88 Joules

Using the final kinetic energy = 1/2 mv²

3981.88 = 1/2 × 42 × v²

3981.88 = 21 v²

v² = 3981.88/21

v² =189.61

v = $\sqrt{189.61}$

v = 13.78 m/s

Therefore, the speed at the bottom = 13.78 m/s

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