Solution to problem 7.6.7 from the collection of Kepe O.E.

Task 7.6.7:

For a point moving at a speed given by the formulas $v_x = 0.2 t^2$, $v_y = 3$ m/s, it is necessary to determine its tangential acceleration at time $t=2.5$ s.

Answer:

The tangential acceleration of a point at time $t$ is determined by the formula:

$a_\text{т} = \sqrt{\left(\dfrac{dv_x}{dt}\right)^2 + \left(\dfrac{dv_y}{dt}\right)^2}$

Finding the derivatives of speeds:

$\dfrac{dv_x}{dt} = 0.4t$ m/s$^2$

$\dfrac{dv_y}{dt} = 0$ м/с$^2$

We substitute the values ​​of the derivatives and the time $t=2.5$ s into the formula for tangential acceleration:

$a_\text{т} = \sqrt{\left(0,4 \cdot 2,5\right)^2 + 0^2} = \sqrt{1} = 1$ м/с$^2$

Answer: $a_\text{t} = 1$ m/s$^2$.

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Solution to problem 7.6.7 from the collection of Kepe O.?. associated with the determination of tangential acceleration at time t = 2.5 s. It is given that the projection of velocity along the x axis is equal to 0.2t^2, and the projection of velocity along the y axis is equal to 3 m/s. It is necessary to find the tangential acceleration at time t = 2.5 s.

To solve the problem, it is necessary to find the derivative of the velocity projection along the x axis with respect to time, and then substitute the time value t = 2.5 s into the resulting expression. The resulting value will be the tangential acceleration at the specified time.

The expression for the velocity projection along the x axis can be written as vx = 0.2t^2. Let's find the derivative of this expression with respect to time t:

dvx/dt = d/dt (0,2t^2) = 0,4t

Let's substitute the time value t = 2.5 s:

dvx/dt |t=2,5 = 0,4 x 2,5 = 1

Thus, the tangential acceleration at time t = 2.5 s is equal to 1 m/s^2. Answer: 1 m/s^2 = 0.385.

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