Chapter 9: First-Order Differential Equations - Section 9.3 - Applications - Exercises 9.3 - Page 544: 4

$$s(t)=1.32(1-e^{-0.606t})$$

We have: $s(t)=\dfrac{v_0 m}{k}(1-e^{-kt/m} )$ Since, $\dfrac{v_0 m}{k}=1.32$ Plug in the given data; we have: $\dfrac{(0.80)(49.90)}{k}=1.32$ $\implies k \approx 30.2424$ Now, $$s(t)=\dfrac{v_0 m}{k}(1-e^{-kt/m} )=1.32(1-e^{-(30)t/49.90} )$$ So, $$s(t)=1.32(1-e^{-0.606t})$$