A particle moving in the *xy*-plane has velocity $\vec{v}_0=v_{0x} \ \hat{i} + v_{0y} \ \hat{j}$ at $t=0$. It undergoes acceleration $\vec{a} = bt \ \hat{i} - c v_{y} \ \hat{j}$ , where b and c are constants. Find an expression for the particle's velocity at a later time $t$. **Given** at $t=0$ $\vec{v}_0=v_{0x} \ \hat{i} + v_{0y} \ \hat{j} \tag{a}$ **Unknown** at some time $t$ $\vec{v}(t) = v_x \ \hat{i} + {v_y} \ \hat{j} + v_{0x} \ \hat{i} + v_{0y} \ \hat{j} \tag{1}$ **Strategy** Focus on the components then combine them to the expression for the particle's velocity. **Vertical Component ($v_y \ \hat{j}$)** $a_y = -c \ v_y \tag{2}$ By definition $a_y = \frac{dv_y}{dt} \tag{3}$ Substitute the definition into the equation 2. $ \frac{dv_y}{dt} = -c \ v_y \tag{4}$ Rearrange equation 4 assuming $v_y \not= 0$ $ \frac{1}{v_y} \cdot d{v_y} = -c \ dt \tag{5}$ Now integrate both sides, and pay attention to the lower and upper limit of integration. $\int ^{v_y}_{v_{0y}}{\frac{1}{v_y} \cdot d{v_y}} = -c \int ^{t}_{0} { \ dt} \tag{6}$ The integral of $\frac{1}{v_y}$ is the natural logarithm of $v_y$ and the integral of $-c$ is $-ct$ plus a constant $C$ (the $C$ is 0 when bounds are applied) . ${ln |{v_y}|} \Biggr|_{v_{0y}}^{v_y}= {-ct} \Biggr|_0^t \tag{6}$ $\ln|{v_y}| - \ln|{v_{0y}|} = -ct \tag{7}$ $\ln\Biggr(\frac{|v_y|}{|v_{0y}|} \Biggr) = -ct \tag{8}$ Use the rule that $\ln x = y$ is $x = e^y$ $\frac{|v_y|}{|v_{0y}|} = e^{-ct} \tag{9}$ Move the $v_{0y}$ to the other side. ${|v_y|} = v_{0y} \ e^{-ct} \tag{10}$ **Horizontal Component ($v_x \ \hat{i}$)** $a_x = b \ t \ v_x \tag{11}$ $\int ^{v_{x}}_{v_{0x}} a_x = \int ^t_0 b \ t \ v_x \tag{12}$ $v_x(t) \Biggr|^{v_{x}}_{v_{0x}} = \frac{1}{2}bt^2 \ v_x \Biggr|^t_0 \tag{13}$ $v_x - v_{0x} = \frac{1}{2}bt^2 \tag{14}$ Move the $v_{0x}$ to the other side. $v_x(t) = \frac{1}{2} bt^2 +v_{0x} \tag{15}$ **Combine Vertical and Horizontal** Remember to add the initial velocities for each component when using the bounds. $v(t) = (\frac{1}{2} bt^2 +v_{0x}) \ \hat{i} + (v_{0y} \ e^{-ct}) \ \hat{j} \tag{16}$