Question

An object initially at rest experiences an acceleration of 0.565 m/s2 to the South for a time of 4.02 seconds. It then increases its acceleration to 1.25 m/s2 to the South for an additional 3.19 seconds. What is the total distance moved by the object, in units of meters?

Answer 1

${\huge{\red{\underline{Answer:}}}}$

Distance travelled (S) = 18.16 m

${\huge{\blue{\underline{Given:}}}}$

initial velocity (u) = 0 m/s

acceleration (${\bold{a_1}}$) = 0.565 m/s²

time (${\bold{t_1}}$) = 4.02 sec

acceleration (${\bold{a_2}}$) = 1.25 m/s²

time (${\bold{t_2}}$) = 3.19 sec

${\huge{\orange{\underline{Formulas:}}}}$

S = ut + 1/2 at²

v = u + at

${\huge{\green{\underline{Solution:}}}}$

${\bold{S_1}}$ = u ${\bold{t_1}}$ + 1/2 ${\bold{a_1}}$${\bold{t_1}}$²

= (0)(4.02) + 1/2 (0.565)(4.02)²

= 0.565 × 4.02 × 2.01 = 4.56 m

v = u + ${\bold{a_1}}$ ${\bold{t_1}}$

= 0 + (0.565)(4.02) = 2.27 m/s

________________________________

${\bold{S_2}}$ = v ${\bold{t_2}}$ + 1/2 ${\bold{a_2}}$${\bold{t_2}}$²

= (2.27)(3.19) + 1/2 (1.25)(3.19)²

= 7.24 + 6.36 = 13.6 m

________________________________

${\bold{S}}$= ${\bold{S_1}}$+ ${\bold{S_2}}$

= 4.56 + 13.6

= ${\bold{\underline{18.16}}}$

Answer 2

18.16 m

$<marquee direction="left" > Give thanks and follow me </marquee>$