Question

Starting form rest a particle is first accelerated for time $t_1$ with constant acceleration $a_1$ and then stops in time $t_2$ with constant retardation $a_2$. Let $v_1$ be the average velocity in this case and $s_1$ be the total displacement. In the second case, it is accelerated for the same time $t_1$ with constant acceleration $2a_1$ and comes to rest with constant retardation $a_2$ in time $t_3$. If $v_2$ is the average velocity in this case and $s_2$ the total displacement. Then

• A. $v_2=2v_1$
• B. $2v_1<v_2<4v_1$
• C. $s_2=2s_1$
• D. $2s_1<s_2<4s_1$

Answer 1

Answer: (A), (D)

The correct options are
A $v_2=2v_1$
D $2s_1<s_2<4s_1$

Let us take $v-t$ curve here, as it gives acceleration as well as displacement.
For case $1:$
So, we have $a_1=\frac{V_{max}}{t_1}$ and $a_2=\frac{V_{max}}{t_2}$
from above two we can say that $V_{max}=a_1{t}_1=a_2{t}_2$.....(i)
Also, $s_1=$ Area under the curve $=\frac{1}{2}\times V_{max}\times (t_1+t_2)$
As average velocity for this case is given as $v_1$
$\Rightarrow v_1=\frac{s_1}{t_1+t_2}=\frac{\frac{1}{2}\times V_{max}\times (t_1+t_2)}{(t_1+t_2)}$
$\Rightarrow v_1=\frac{V_{max}}{2}=\frac{a_1{t}_1}{2}=\frac{a_2{t}_2}{2}$....(ii)

For case $2:$
So, we have $2a_1=\frac{V_{max}^{{}^{\prime }}}{t_1}$ and $a_2=\frac{V_{max}^{{}^{\prime }}}{t_3}$
from above two we can say that $V_{max}^{{}^{\prime }}=2a_1{t}_1=a_2{t}_3$....(iii)
On putting the value of $a_2=\frac{a_1{t}_1}{t_2}$ from eq (i) into eq (iii) we get
$t_3=2t_2$....(iv)
Also, $s_2=$ Area under the curve $=\frac{1}{2}\times V_{max}^{{}^{\prime }}\times (t_1+t_3)$
As average velocity for this case is given as $v_2$
$\Rightarrow v_2=\frac{s_2}{t_1+t_3}=\frac{\frac{1}{2}\times V_{max}^{{}^{\prime }}\times (t_1+t_3)}{(t_1+t_3)}$
$\Rightarrow v_2=\frac{V_{max}^{{}^{\prime }}}{2}$
$\Rightarrow v_2=\frac{2a_1{t}_1}{2}=a_1{t}_1$
Also, $v_2=\frac{a_2{t}_3}{2}=\frac{2a_2{t}_2}{2}=a_2{t}_2$
So, combindly we can write it as
$v_2=\frac{V_{max}^{{}^{\prime }}}{2}=a_1{t}_1=a_2{t}_2$.....(v)
Now, from eq (ii) and (v), we can say that $v_2=2v_1$

Also, checking for other relation we have
$s_1=\frac{1}{2}\times V_{max}\times (t_1+t_2)=\frac{1}{2}\times a_1{t}_1\times (t_1+t_2)$
or, $2s_1=a_1{t}_1(t_1+t_2)$
and, $s_2=\frac{1}{2}\times V_{max}^{{}^{\prime }}\times (t_1+t_3)=\frac{1}{2}\times 2a_1{t}_1\times (t_1+t_3)$
or, $s_2=a_1{t}_1(t_1+2t_2)$
Therefore, from above two findings we can say that $2s_1\ne s_2$ rather, $2s_1<s_2$ as the value of $s_2$ is greater by $t_2$ compared to $2s_1$.
Also, $4s_1=2\times 2s_1=2a_1{t}_1(t_1+t_2)>s_2$
Thus, we can say that $2s_1<s_2<4s_1$