Question

A car accelerated from 0.0 m/s to 24 m/s at a constant rate in 6.0 s. How far did the car travel between t= 3.0 s and t= 6.0 s ?

Answer 1

Answer on Question #46823, Physics, Mechanics | Kinematics | Dynamics

Since the car is moving with constant acceleration and initial velocity is zero, the time dependence of velocity with respect to time is $v(t) = at$ . Using the fact, that velocity is $24\frac{m}{s}$ at $t = 6$ ,

obtain $a = \frac{24\frac{m}{s}}{6s} = 4\frac{m}{s^2}$

Hence, the law of motion is $S(t) = \frac{at^2}{2} = 2t^2$ .

Thus, the car traveled $l = S(t = 6) - S(t = 3) = 2(6^2 - 3^2) = 54m$ between $t = 3s$ and $t = 6s$ .

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