Question

A car accelerates 3.0 m/s from rest. After 4.0 s, (a) what is its velocity? (b) what distance is covered

Answer 1

Question 24639

According to given conditions, $a = 3\frac{m}{s^2}; t = 4s; v_0 = 0\frac{m}{s}$ .

For accelerated motion, velocity as function of time is $v(t) = v_0 + at$ , and covered distance is $S(t) = \int_0^t v(t)dt = v_0t + \frac{at^2}{2}$ .

Hence, for our task velocity after 4 seconds is $v = v_{0} + at = 0 + 3\frac{m}{s^{2}} \cdot 4s = 12\frac{m}{s}$ .

Covered distance after 4 seconds is $S = v_{0}t + \frac{at^{2}}{2} = 0 + \frac{3\cdot 4^{2}}{2} = 24m$ .