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3 years ago

Cars A and B are racing each other along the same straight road in the following manner:

Physics

1 answer:

zysi [14]3 years ago

8 0

Answer:

$\frac{x_{o}}{v_B-V_A} =t$

Explanation:

Represent the car's position as a function

$x_o=$ "head start"

$x_{A}(t) = x_{o} + v_{A}t\\x_B(t)=v_Bt \\$

$Remember: v_B>v_A$

"cathching up means" that $x_A(t)=x_B(t)$

$x_{o} + v_{A}t =v_Bt\\x_{o} = v_Bt -v_{A}t\\x_{o} = t(v_B-V_A)\\\frac{x_{o}}{v_B-V_A} =t, where \ v_B>v_A$

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An airplane starts from rest and accelerates at 10.8 m/s2 . what is its speed at the end of a 400 m long runway?

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The final speed of an airplane is v = 92.95 m/s

The rate of change of position of an object in any direction is known as speed i.e. in other word, Speed is measured as the ratio of distance to the time in which the distance was covered.

Solution-

Here given,

Acceleration a= 10.8 m/s2 .

Displacement (s)= 400m

Then to find final speed of airplane v=?

Therefore from equation of motion can be written as,

v²=u²+ 2as

where, u is initial speed, v is final speed ,a is acceleration and s is displacement of the airplane. Therefore by putting the value of a & s in above equation and (u =0) i.e. the initial speed of airplane is zero.

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v²=8640m/s

v=92.95m/s

hence the final speed of airplane v =92.95m/s

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Answer:

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Explanation:

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A certain spring stores 10.0J of potential energy when it isstretched by 2.00cm from its equilibrium position.How much potential

qaws [65]

Answer:

Answered

Explanation:

x= 0.02 m

E_p= 10.0 J

E_p= 0.5kx^2

10= 0.5k(0.02)^2

solving we get

K= 50.0 N/m

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