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3 years ago

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A cyclist traveling at 30.0m/s along a straight road comes uniformly to stop in 5.00s. Determine the stopping acceleration, the

stopping distance.

1 answer:

Amiraneli [1.4K]3 years ago

7 0

Answer:

I. Stopping acceleration = 6 m/s²

II. Stopping distance, S = 75 meters

Explanation:

Given the following data;

Final velocity = 30 m/s

Time = 5 seconds

To find the stopping acceleration;

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation;

$a = \frac{30 - 0}{5}$

$a = \frac{30}{5}$

Acceleration = 6 m/s²

II. To find the stopping distance, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

30² = 0² + 2*6*S

900 = 12S

S = 900/12

S = 75 meters

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Answer:

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Answer:

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We need to calculate the speed at time, t = 7.0 * $10^{-8}$ s.

We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed $v_{x}$ on the axis.

The electric force acting on the proton moves in the y direction, so this means it is moving with velocity $v_{y}$ in the y axis.

Hence, the resultant velocity of the proton is given by:

$v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v_{x}$ = 990m/s from the question. We need to find $v_{y}$ and then the resultant velocity v.

Electric field is given in terms of surface charge density, σ as:

E = σ/ε0

where ε0 = permittivity of free space

=> $E = \frac{2.34 * 10^{-9} } {2 * 8.85418782 * 10^{-12} }$

E = 132 N/C

Electric Force, F is given in terms of Electric field:

F = eE

where e = electronic charge

=> F = ma = eE

∴ a = eE/m

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$a = \frac{1.60 * 10^{-19} * 132}{1.672 * 10^{-27} }$

a = 1.3 * $10^{10}$ m/ $s^{2}$

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$a = \frac{v_{y} - v_{0}}{t} \\\\=> v_{y} = v_{0} + at$

$v_{y}$ = 0 + (1.3 * $10^{10}$ * 7.0 * $10^{-8}$ )

$v_{y}$ = 910 m/s

∴ $v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v = \sqrt{990^{2} + 910^{2} }$

$v = \sqrt{1808200}$

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