Question

The velocity of a vehicle was determined at different points. If the first velocity is determined to be 100 when the distance covered is 105m at 210m the velocity has increased to 120m/s continue on that journey the velocity increase to 138 when the distance is 250 at point when velocity is 142 the distance covered is 289km.

Finally the velocity increase to 160 when the distance covered is 310m.

Determine the acceleration of the vehicle graphically using second equation of motion

Answer 1

As per the given question,

displacement between $(d)=210-105=105m$

Initial velocity $u=100 m/s$

final velocity$v=120 m/s$

$a=\frac{v^2-u^2}{2d}$

$\Rightarrow a_1 = \frac{120^2-100^2}{2\times 105}$

$\Rightarrow a_1 = \frac{4400}{210}=20.95m/s^2$

For second case,

$a_1=\frac{138^2-120^2}{2\times 40}=58.05m/s^2$

for third case,

$a_2=\frac{142^2-138^2}{2\times 39}=14.35 m/s^2$

for the fourth case,

$a_3=\frac{160^2-142^2}{2\times 21}m/s^2$

$=129.42 m/s^2$