Question

17. A car starts moving along a line, first with an acceleration a=5 ms 2 starting from rest, then uniformly and
finally decelerating at the same rate, comes to rest in the total time of 25 seconds (ty), then average velocity
during the time is equal to v=72 kmph. How long does the particle move uniformly?
A) 25 seconds
B) 2.5 seconds C)1.5 hours
D) 15 seconds
​

Answer 1

Answer:-

$\Large\boxed{\sf{15\ seconds}}$

Solution:-

Distance travelled with the acceleration $\sf{a=5\ m\ s^{-2}}$ starting from rest is,

$\longrightarrow\sf{s_1=(0)t_1+\dfrac{1}{2}\cdot5(t_1)^2}$

$\longrightarrow\sf{s_1=\dfrac{5}{2}(t_1)^2}$

Speed attained after $\sf{t_1}$ seconds is,

$\longrightarrow\sf{v_1=0+5t_1}$

$\longrightarrow\sf{v_1=5t_1}$

Distance travelled uniformly is,

$\longrightarrow\sf{s_2=5t_1t_2+\dfrac{1}{2}(0)(t_2)^2}$

$\longrightarrow\sf{s_2=5t_1t_2}$

Speed attained after travelling uniformly (i.e., travelling after $\sf{t_1+t_2}$ seconds of its journey from rest) is,

$\longrightarrow\sf{v_2=5t_1+(0)t_2}$

$\longrightarrow\sf{v_2=5t_1}$

Distance travelled with the retardation $\sf{a=-5\ m\ s^{-2}}$ is,

$\longrightarrow\sf{s_3=5t_1t_3+\dfrac{1}{2}\cdot(-5)(t_3)^2}$

$\longrightarrow\sf{s_3=5t_1t_3-\dfrac{5}{2}(t_3)^2\quad\quad\dots(1)}$

Velocity attained when being brought to rest is 0, i.e.,

$\longrightarrow\sf{0=5t_1+(-5)t_3}$

$\longrightarrow\sf{0=5t_1-5t_3}$

$\longrightarrow\sf{t_1=t_3}$

Then (1) becomes,

$\longrightarrow\sf{s_3=5(t_1)^2-\dfrac{5}{2}(t_1)^2}$

$\longrightarrow\sf{s_3=\dfrac{5}{2}(t_1)^2}$

$\longrightarrow\sf{s_3=s_1\quad\quad\dots(2)}$

Total time,

$\longrightarrow\sf{t_1+t_2+t_3=25}$

$\longrightarrow\sf{2t_1+t_2=25}$

$\longrightarrow\sf{t_1=\dfrac{25-t_2}{2}\quad\quad\dots(3)}$

$\longrightarrow\sf{t_1+t_2=\dfrac{25-t_2}{2}+t_2}$

$\longrightarrow\sf{t_1+t_2=\dfrac{25+t_2}{2}\quad\quad\dots(4)}$

Total distance travelled $\sf{(72\ km\ h^{-1}=20\ m\ s^{-1}),}$

$\longrightarrow\sf{s_1+s_2+s_3=Total\ time\cdot Average\ Velocity}$

From (2),

$\longrightarrow\sf{2s_1+s_2=25\times20}$

$\longrightarrow\sf{5(t_1)^2+5t_1t_2=500}$

$\longrightarrow\sf{(t_1)^2+t_1t_2=100}$

$\longrightarrow\sf{t_1[t_1+t_2]=100}$

From (3) and (4),

$\longrightarrow\sf{\dfrac{25-t_2}{2}\cdot\dfrac{25+t_2}{2}=100}$

$\longrightarrow\sf{\dfrac{(25-t_2)(25+t_2)}{4}=100}$

$\longrightarrow\sf{625-(t_2)^2=400}$

$\longrightarrow\sf{(t_2)^2=225}$

$\longrightarrow\sf{\underline{\underline{t_2=15\ s}}}$

Thus the car moves with uniform speed for 15 seconds.