Question

a bus acceleration from rest a constant rate alpha for some time after which it decelerates at a constant raye beta to come to rest .if the total time elapsed is t seconds ,then the total distance travelled is .. answer please
answer in right way step by step

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Answer 1

Answer:

the total distance is given as follows

Answer 2

The distance travelled with the constant acceleration $\sf{\alpha}$ from rest for a time $\sf{t_1}$ is,

$\longrightarrow\sf{s_1=(0)t_1+\dfrac{1}{2}\alpha(t_1)^2}$

$\longrightarrow\sf{s_1=\dfrac{1}{2}\alpha(t_1)^2}$

The velocity attained after $\sf{t_1}$ seconds will be,

$\longrightarrow\sf{v_1=0+\alpha t_1}$

$\longrightarrow\sf{v_1=\alpha t_1}$

The distance travelled with the constant deceleration $\sf{-\beta}$ for a time $\sf{t-t_1}$ seconds is (total time = $\sf{t}$)

$\longrightarrow\sf{s_2=v_1(t-t_1)-\dfrac{1}{2}\beta(t-t_1)^2}$

$\longrightarrow\sf{s_2=\alpha t_1(t-t_1)-\dfrac{1}{2}\beta(t-t_1)^2}$

$\longrightarrow\sf{s_2=\alpha t_1t-\alpha(t_1)^2-\dfrac{1}{2}\beta(t^2-2t_1t+(t_1)^2)}$

$\longrightarrow\sf{s_2=\alpha t_1t-\alpha(t_1)^2-\dfrac{1}{2}\beta t^2+\beta t_1t-\dfrac{1}{2}\beta(t_1)^2}$

Since the bus is coming to rest, final velocity is zero.

$\longrightarrow\sf{0=\alpha t_1-\beta(t-t_1)}$

$\longrightarrow\sf{\alpha t_1=\beta t-\beta t_1}$

$\longrightarrow\sf{t_1=\dfrac{\beta}{\alpha+\beta}\ t\quad\quad\dots(1)}$

Then total displacement,

$\longrightarrow\sf{s=s_1+s_2}$

$\longrightarrow\sf{s=\dfrac{1}{2}\alpha(t_1)^2+\alpha t_1t-\alpha(t_1)^2-\dfrac{1}{2}\beta t^2+\beta t_1t-\dfrac{1}{2}\beta(t_1)^2}$

$\longrightarrow\sf{s=-\dfrac{1}{2}\left(\alpha+\beta\right)(t_1)^2+(\alpha+\beta)t_1t-\dfrac{1}{2}\beta t^2}$

From (1),

$\longrightarrow\sf{s=-\dfrac{1}{2}\left(\alpha+\beta\right)\left(\dfrac{\beta}{\alpha+\beta}\ t\right)^2+(\alpha+\beta)\left(\dfrac{\beta}{\alpha+\beta}\ t\right)t-\dfrac{1}{2}\beta t^2}$

$\longrightarrow\sf{s=-\dfrac{\beta^2t}{2(\alpha+\beta)}+\beta t^2-\dfrac{1}{2}\beta t^2}$

$\longrightarrow\sf{\underline{\underline{s=\dfrac{\beta t^2(\alpha+\beta-1)}{2(\alpha+\beta)}}}}$

If deceleration is taken as $\sf{\beta,}$

$\longrightarrow\sf{\underline{\underline{s=-\dfrac{\beta t^2(\alpha-\beta-1)}{2(\alpha-\beta)}}}}$