Question

a 1800kg car accelerates over a distance of 120meter due to a net force of 4200newton id its initial speed was 40km per hour what is its final speed?

Answer 1

Question 18991

According to 2nd Newton's law, the acceleration is $a = \frac{F}{m} = \frac{7}{3} m / s^2$ . From the given distance, solve for time: $S = v_0 t + \frac{a t^2}{2} = 120$ ; $120 = \frac{100}{9} t + \frac{7}{6} t^2 \Rightarrow t \approx 6.44 \, s$ (Here we converted the initial velocity from km/s into m/s). Plugging this time into the law of change of velocity $v = v_0 + at$ , obtain $v = \frac{100}{9} + \frac{7}{6} \cdot 6.44 \approx 18.62 \, m / s \approx 67 \, km / h$ .