Class 11 NCERT PHYSICS • Chapter 3: Motion in a Plane • Question 18

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The kinematic equations for constant acceleration in two dimensions are:

\begin{aligned} \vec{r}(t) &= \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2 \\ \vec{v}(t) &= \vec{v}_0 + \vec{a} t \end{aligned}

The speed is the magnitude of the velocity vector:

v = \sqrt{v_x^2 + v_y^2}

Initial position: $\vec{r}_0 = (0 \hat{\mathbf{i}} + 0 \hat{\mathbf{j}})\,\text{m}$
Initial velocity: $\vec{v}_0 = (0 \hat{\mathbf{i}} + 10 \hat{\mathbf{j}})\,\text{m/s}$
Acceleration: $\vec{a} = (8.0 \hat{\mathbf{i}} + 2.0 \hat{\mathbf{j}})\,\text{m/s}^2$
Target x-coordinate: $x = 16\,\text{m}$

Using the x-component of the displacement:

\begin{aligned} x(t) &= v_{0x} t + \frac{1}{2} a_x t^2 \\ 16 &= 0 \cdot t + \frac{1}{2} (8.0) t^2 \\ 16 &= 4.0 t^2 \\ t^2 &= 4.0 \\ t &= 2.0\,\text{s} \end{aligned}

Using the y-component of the displacement:

\begin{aligned} y(t) &= v_{0y} t + \frac{1}{2} a_y t^2 \\ y(2.0) &= 10(2.0) + \frac{1}{2} (2.0) (2.0)^2 \\ &= 20 + 4.0 \\ &= 24.0\,\text{m} \end{aligned}

First, find the velocity components:

\begin{aligned} v_x &= v_{0x} + a_x t \\ &= 0 + 8.0(2.0) = 16.0\,\text{m/s} \\ v_y &= v_{0y} + a_y t \\ &= 10 + 2.0(2.0) = 14.0\,\text{m/s} \end{aligned}

Now, calculate the speed:

\begin{aligned} v &= \sqrt{v_x^2 + v_y^2} \\ &= \sqrt{16.0^2 + 14.0^2} \\ &= \sqrt{256 + 196} \\ &= \sqrt{452} \\ &\approx 21.26\,\text{m/s} \end{aligned}

(a) Time:

2.0\,\text{s}

y-coordinate:

24.0\,\text{m}

(b) Speed:

21.26\,\text{m/s}

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