## AJTurnbull Group Title a car is traveling in a straight line and decelerating at a constant rate. after 3.2 s it traveled 88m and slowed to a speed of 17m/s what was the initial speed of the car? 2 years ago 2 years ago

1. nikvist Group Title

$t=3.2s\quad,\quad s=88m\quad,\quad v=17m/s$$v=v_0-at\quad\Rightarrow\quad a=\frac{v_0-v}{t}$$v^2=v_0^2-2as=v_0^2-2\frac{v_0-v}{t}s=v_0^2-\frac{2s}{t}v_0+\frac{2s}{t}v$$v_0^2-\frac{2s}{t}v_0+\frac{2s}{t}v-v^2=0$$v_0=\frac{\frac{2s}{t}\pm\sqrt{\left(\frac{2s}{t}\right)^2-4\left(\frac{2s}{t}v-v^2\right)}}{2}=$$=\frac{s}{t}\pm\sqrt{\left(\frac{s}{t}\right)^2-\frac{2s}{t}v+v^2}=\frac{s}{t}\pm\sqrt{\left(\frac{s}{t}-v\right)^2}=$$=\frac{s}{t}\pm\left(\frac{s}{t}-v\right)\quad\mbox{(only sign +)}$$v_0=\frac{2s}{t}-v=38m/s$

2. shubham Group Title

OR Use this equation of motion S = {(u+v)/2}t 88m = {(u+17)/2}3.2 u=55-13 = 38m/s

3. AJTurnbull Group Title

how did you get a?

4. shubham Group Title

For constant accelarations, velocity is linear function of time and hence you can assume mean of initial and final velocities to be the average velocity. So, you don't need a actually.

5. AJTurnbull Group Title

oh that makes it easy been trying to figure that out all night