Question

a body starts from rest and accelerates with 2 metre per second square for 10 seconds it then retardes 1 metre per second square and comes to rest find the total displacement covered by the body total time taken and maximum velocity of the body​

Answer 1

$</p><p></p><p>u = 0 \: m/s \\ a_{1} = 2 \: m/{s}^{2} \\ a_{2} = -1 \: m/{s}^{2} \\ t_{1} = 10 \\ t_{2} = ? \\ s = ?</p><p></p><p>$

We know that,

$\boxed{v = u + at}$

$\Rightarrow \qquad v = u + at \\ \Rightarrow \qquad v = 0 + 2 \cdot 10 \\ \Rightarrow \qquad v = 20 \: m/s$

We have, v = 20 m/s

Again,

$\boxed{h = ut + \frac{1}{2}at^2}$

$</p><p>h = ut + \frac{1}{2} at^{2} \\ \Rightarrow \qquad 0 \cdot 10 + \frac{1}{2} \cdot 2 \cdot 10^2 \\ \Rightarrow \qquad \frac{1}{2} \cdot 2 \cdot 100 \\ \Rightarrow \qquad 100 \: m </p><p>$

We have, h = 100 m

Again,

$\boxed{2as = v^2 - u^2}$

$\Rightarrow \qquad 2 \cdot -1 \cdot s = 0 - 20^2$ [ The car comes at rest therefore it's final velocity will be 0 and initial velocity will be v in this case. ]

$\Rightarrow \qquad -2s = -400 \\ \Rightarrow \qquad s = 200$

We have, s = 200 m

Again,

$\boxed{v = u + at_{2}}$

$0 = 20 + (-1 \cdot t_{2}) \\ 0 = 20 - t_{2} \\ t_{2} = 20$

We have $\bold{t_{2} = 20 \: s}$

Maximum velocity = v

=> 20 m/s

Total Time taken = $t_{1} + t_{2}$

=> 10 + 20

=> 30 s

Total Displacement = $s + h$

=> 200 + 100

=> 300 m