Question

what is the value of an acceleration of a ball thrown upwad? How much time does it take for a camel to accelerate 10 m/s from Test to speed of 30?​

Answer 1

Answer:

Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocity function is acceleration,

\[\frac{d}{dt}v(t)=a(t),\]

we can take the indefinite integral of both sides, finding

\[\int \frac{d}{dt}v(t)dt=\int a(t)dt+{C}_{1},\]

where C1 is a constant of integration. Since

\[\int \frac{d}{dt}v(t)dt=v(t)\]

, the velocity is given by

\[v(t)=\int a(t)dt+{C}_{1}.\]

Similarly, the time derivative of the position function is the velocity function,

\[\frac{d}{dt}x(t)=v(t).\]

Thus, we can use the same mathematical manipulations we just used and find

\[x(t)=\int v(t)dt+{C}_{2},\]

where C2 is a second constant of integration.

We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doing the integration in (Figure), we find

\[v(t)=\int adt+{C}_{1}=at+{C}_{1}.\]

If the initial velocity is v(0) = v0, then

\[{v}_{0}=0+{C}_{1}.\]

Then, C1 = v0 and

\[v(t)={v}_{0}+at,\]

which is (Equation). Substituting this expression into (Figure) gives

\[x(t)=\int ({v}_{0}+at)dt+{C}_{2}.\]

Doing the integration, we find

\[x(t)={v}_{0}t+\frac{1}{2}a{t}^{2}+{C}_{2}.\]

If x(0) = x0, we have

\[{x}_{0}=0+0+{C}_{2};\]

so, C2 = x0. Substituting back into the equation for x(t), we finally have

\[x(t)={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2},\]

which is (Equation).

Answer 2

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what is the value of an acceleration of a ball thrown upwad? How much time does it take for a camel to accelerate 10 m/s from Test to speed of 30?

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If you throw the ball upward with a speed of 9.8 m/s, the velocity has a magnitude of 9.8 m/s in the upward direction. Now the ball is under the influence of gravity, which, on the surface of the Earth, causes all free-falling objects to undergo a vertical acceleration of –9.8 m/s2.

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The change in velocity is v – u, which is 30 – 10 = 20 m/s.

The acceleration is the change in velocity ÷ time, which is 20 m/s ÷ 4 s = 5 m/s2.

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hope it's helpful!