A car that accelerates uniformly passes at point A with a speed of 30 kph. It reaches point B, 21 km from A, after 18 minutes. Find the acceleration of the car. A. 0.0206 m/s2 C. 0.126 m/s2 B. 0.0562 m/s2 D. 3.42 m/s2 Solution Click here to… Jhun Vert Wed, 07/03/2024 - 22:27 Solution Click here to expand or collapse this section $s = v_o t + \frac{1}{2}a t^2$ $21 = 30 \left( \frac{18}{60} \right) + \frac{1}{2}a\left( \frac{18}{60} \right)^2$ $\frac{9}{200}a = 12$ $a = \frac{800}{3} ~ \text{km/hr}^2$ $a = \dfrac{800}{3} ~ \dfrac{\text{km}}{\text{hr}^2} \times \dfrac{1000 ~ \text{m}}{1 ~ \text{km}} \times \left( \dfrac{1 ~ \text{hr}}{3600 ~ \text{sec}} \right)^2$ $a = \frac{5}{243} ~ \text{m/sec}^2 = 0.02057 ~ \text{m/sec}^2$ Log in or register to post comments Log in or register to post comments
Solution Click here to… Jhun Vert Wed, 07/03/2024 - 22:27 Solution Click here to expand or collapse this section $s = v_o t + \frac{1}{2}a t^2$ $21 = 30 \left( \frac{18}{60} \right) + \frac{1}{2}a\left( \frac{18}{60} \right)^2$ $\frac{9}{200}a = 12$ $a = \frac{800}{3} ~ \text{km/hr}^2$ $a = \dfrac{800}{3} ~ \dfrac{\text{km}}{\text{hr}^2} \times \dfrac{1000 ~ \text{m}}{1 ~ \text{km}} \times \left( \dfrac{1 ~ \text{hr}}{3600 ~ \text{sec}} \right)^2$ $a = \frac{5}{243} ~ \text{m/sec}^2 = 0.02057 ~ \text{m/sec}^2$ Log in or register to post comments
Solution Click here to…