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wel
3 years ago
9

Vinny is on a motorcycle at rest, 200 m away from a ramp that jumps over a gully. Calculate the minimum constant acceleration Vi

nny must have to get to the ramp in 8 s before his pursuers catch up with him.
Physics
1 answer:
soldier1979 [14.2K]3 years ago
7 0

Answer: 6.25 m/s^2

Explanation:

The distance between Vinny and the ramp is 200m

And he has 8 seconds (At max) to reach that distance.

The initial velocity is 0m/s

The initial position is 0m

Now, we want to find the constant acceleration in order to do this, so suppose that we have a constant acceleration A.

a(t) = A.

To have the velocity, we must integrate over time, and remember that the constant of integration is equal to zero because the initial velocity is zero.

v(t) = A*t

For the position, we integrate again over time.

p(t) = 0.5*A*t^2

And we want to travel 200m in 8 seconds, then:

p(8s) = 200m

0.5*A*(8s)^2 = 200m

A*32s^2 = 200m

A = 200m/32s^2 = 6.25 m/s^2

This is the minimum acceleration in order to do this, if Vinny has a larger acceleration he will travel the 200m in a smaller time.

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Answer:

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6 0
3 years ago
A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i
tresset_1 [31]

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

5 0
3 years ago
A biologist looking through a microscope sees a bacterium at r⃗ 1=2.2i^+3.7j^−1.2k^μm(1μm=10−6m). After 6.2 s , it's at r⃗ 2=4.6
stiv31 [10]

The Average velocity for the bacterium is 0.75 unit/sec.

<u>Explanation:</u>

The given values are in the vector form

Where,

dS = distance covered

dT = time interval

Now, to calculate distance covered, we have

|d S|=\sqrt{d S^{2}}

&

d S=r_{2}-r_{1}

d S=(4.6 i+1.9 k)-(2.2 i+3.7 j - 1.2 k)

d S=(4.6-2.2) i+(0-3.7) j+(1.9+1.2) k

d S=2.4 i-3.7 j+3.1 k

Now, putting these values in the standard formula to evaluate the average velocity, we get;

v_{a v g}=\frac{|\mathrm{d} S|}{d T}

v(a v g)=\frac{|\sqrt{\left\{\left(2.4^{2}\right)+\left(3.7^{2}\right)+\left(3.1^{2}\right)\right\}}|}{7.2}

As dT=7.2 sec

Now,

Solving the equation, we get;

v(a v g)=\frac{5.390732789}{7.2}

\begin{aligned}&v(a v g)=\frac{5.39}{7.2}\\&v(a v g)=0.748611111\\&v(a v g)=0.75 \text { units / sec }\end{aligned}

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3 0
3 years ago
A table tennis ball with a mass of 0.003kg and soccer with a mass of 0.43 kg. are both set in motion at 16 m"s. Calculate and co
Aliun [14]

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7 0
3 years ago
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poizon [28]

Answer: Heyaa! ~

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