Question

A student has been working on problems much like yours and has arrived at the following set of kinematic equations

9.0m= 1/2 (2.0 m/s^2 ) t_1^2

v_1=(2.0 m/s^2 ) t_1

36 m-9.0 m= v_1 (t_2- t_1 )

Work your way backward to find a real problem for which this is the solution.

Answer 1

The givne data is $s=\frac{1}{2}x2xt_{1}^{2}$

   $v_{1}=2xt_{1}$

  

From equation -1, $9=\frac{1}{2}2xt_{1}^{2}$    $\Rightarrow 9=t_{1}^{2}$

   $\Rightarrow t_{1}= 3s$

from equation-2, v₁ =2xt₁

v= 2x3 = 6m/s

from equation-3,

36-9 = 6x(t₂-t₁)

27=6(t₂-3)

4.5=t₂ -3

t₂ = 7.5s

A body stating from rest moves under acceleration 2m/s², is accelerated for 3 seconds, later moves with constant velocity for 4.5 seconds. Find displacement during acceleration, total displacement in 7.5 seconds.