Question

A streetcar moves rectilinear from station A to the next station B (from rest to rest) with an acceleration varying according to the law f = a – bx; where a and b are constants and x is the distance from station A. The distance between the two stations and the maximum velocity are:

1. $x = \dfrac{{2a}}{b}$ , ${v_{\max }} = \dfrac{a}{{\sqrt b }}$ ;
2. $x = \dfrac{b}{{2a}}$,${v_{\max }} = \dfrac{a}{{\sqrt b }}$;
3. $x = \dfrac{a}{{2b}}$,${v_{\max }} = \dfrac{b}{{\sqrt a }}$;
4. $x = \dfrac{a}{b}$, ${v_{\max }} = \dfrac{{\sqrt a }}{b}$;

Answer 1

Here we need to find the velocity at an instant by differentiating the given acceleration. Acceleration is the rate of change of velocity with respect to time. Integrate the instant velocity and short distance together and find out the distance between A and B by keeping the final velocity zero. Use the given formula and apply the condition of maximum velocity and then find out the max velocity by placing it in the found formula for the whole velocity.

Complete step by step solution:
Step 1: First find out the velocity at an instant:
For that we need to find out the acceleration at an instant:
$f = \dfrac{{dv}}{{dt}} \times \dfrac{{dx}}{{dx}}$ ;
$\Rightarrow f = \dfrac{{dx}}{{dt}} \times \dfrac{{dv}}{{dx}}$;
Now, we know that the velocity is change distance upon change in time:
$\Rightarrow f = v\dfrac{{dv}}{{dx}}$;
Now, we have been given the law for acceleration as f = a – bx:
$\Rightarrow a - bx = v\dfrac{{dv}}{{dx}}$;
$\Rightarrow dx\left( {a - bx} \right) = vdv$;
Now, integrate on both sides to get total velocity:
$\Rightarrow \int\limits_0^x {dx\left( {a - bx} \right)} = \int\limits_0^v {vdv}$;
$\Rightarrow \left[ {ax - \dfrac{{b{x^2}}}{2}} \right]_0^x = \left[ {\dfrac{{{v^2}}}{2}} \right]_0^v$;
$\Rightarrow ax - \dfrac{{b{x^2}}}{2} = \dfrac{{{v^2}}}{2}$;
Now, solve for “v” by taking the root on both sides:
$\Rightarrow \sqrt {ax - \dfrac{{b{x^2}}}{2}} = \sqrt {\dfrac{{{v^2}}}{2}}$;
Now, in the question the final velocity is zero:
$\Rightarrow \sqrt {ax - \dfrac{{b{x^2}}}{2}} = 0$;
$\Rightarrow ax - \dfrac{{b{x^2}}}{2} = 0$;
Take the common out:
$\Rightarrow x\left( {a - \dfrac{{bx}}{2}} \right) = 0$;
$\Rightarrow \left( {a - \dfrac{{bx}}{2}} \right) = 0$;
Now, solve for x which is the distance:
$\Rightarrow x = \dfrac{{2a}}{b}$;
Step 2:
Now, we need to find out the maximum velocity and at maximum velocity there will be no change in the acceleration so, acceleration will be constant or zero:
$f = a - bx$ ;
$\Rightarrow 0 = a - bx$;
Solve for x:
$\Rightarrow x = \dfrac{a}{b}$;
Now, we know the formula for velocity “$\sqrt {ax - \dfrac{{b{x^2}}}{2}} = \sqrt {\dfrac{{{v^2}}}{2}}$”
$\sqrt {a\left( {\dfrac{a}{b}} \right) - \dfrac{{b{{\left( {\dfrac{a}{b}} \right)}^2}}}{2}} = \sqrt {\dfrac{{v_{\max }^2}}{2}}$;
$\Rightarrow \sqrt {\left( {\dfrac{{{a^2}}}{b}} \right) - \dfrac{{\left( {\dfrac{{{a^2}}}{b}} \right)}}{2}} = \sqrt {\dfrac{{v_{\max }^2}}{2}}$;
Remove the roots:
$\Rightarrow \left( {\dfrac{{{a^2}}}{b}} \right) - \left( {\dfrac{{{a^2}}}{{2b}}} \right) = \dfrac{{v_{\max }^2}}{2}$;
$\Rightarrow \dfrac{{2{a^2} - {a^2}}}{{2b}} = \dfrac{{v_{\max }^2}}{2}$;
Do, the necessary calculation:
$\Rightarrow \sqrt {\dfrac{{{a^2}}}{b}} = {v_{\max }}$;
The maximum velocity is:
$\Rightarrow {v_{\max }} = \dfrac{a}{{\sqrt b }}$;

Final Answer: Option “1” is correct. Therefore, the distance between the two stations and the maximum velocity are $x = \dfrac{{2a}}{b}$ and ${v_{\max }} = \dfrac{a}{{\sqrt b }}$.

Note: Here, we need to be very careful while solving the question, in the formula f = a – bx; f is not force but acceleration and a is not acceleration but a constant, don’t get confused. Here, we first found out the velocity at the instant and then we integrated it for the whole distance and velocity. The final velocity is zero and at maximum velocity the acceleration would be zero.