Could someone please explain how you get the answer for the following question:
A car of mass 1300 kg is traveling at a speed of 30ms^-1 along a straight horizontal
motorway when the driver sees a traffic jam ahead and applies the brakes for 15 s.
The car covers a distance of 270 m while the driver is braking.
a) Find the speed of the car at the end of the 15 s braking interval.
My Working:
S = 1/2(v-u)t
270=15V-1/2(2)(15)^2
15V=495
V=33
Thanks
From the problem statement let $v_0= 30 \frac{m}{s}$, $m=1300 kg$, $\Delta x=270m$, $\Delta t=15s$, and let the force applied to the breaks be $F_b$. We have, $$ ma=-F_b $$ which can be written, $$ mv\frac{dv}{dx}=-F_b $$ Integrating we get $$ \frac{1}{2}v^2 |_{v_0}^{v_f}=-F_b \Delta x\\ \frac{1}{2}v_f^2=-F_b \Delta x + \frac{1}{2}mv_0^2\\ v_f = \sqrt{\frac{2}{m}(-F_b \Delta x + \frac{1}{2}mv_0^2)} $$ We need to determine $F_b$. Write, $$ m\frac{dv}{dt}=-F_b $$ integrate to get $$ v=\frac{dx}{dt}=\frac{-F_b}{m}t + v_0 $$ Integrate again to get $$ \Delta x=\frac{-F_b}{2m} t^2 + v_0 t\\ F_b=2m\frac{( v_0 t-\Delta x)}{t^2} $$ Substituting $F_b$ in the equation for final velocity, we find, $$ v_f = \sqrt{ v_0^2 - \frac{4\Delta x ( v_0 t -\Delta x)}{t^2}} $$