Question

Calculate the acceleration and distance of the body moving with 5 meter per seconds which comes to rest after travelling for 6 seconds?​

Answer 1

Answer:

acceleration= -0.833m/s²

distance= 15m

Explanation:

given,

u(initial)= 5 m/s

t=6 s v(final)=0m/s

wkt,

v=u+at

a= -u/t (v=0)

a= -5/6

a= -0.833m/s²

wkt,

v²-u²=2as (s-distance travelled)

s= -25/2*(-5/6)

s= 25*6/5*2

s= 15m

Answer 2

Answer:

$v \: = final \: velocity \: = 0m /sec(comes \: at \: rest \: after \: 6sec)$

$u \: = inital \: velocity \: = 5m/sec$

$t = time \: = 6sec$

$a \: = acceleration \:$

$from \: newton \: frist \: equation \\ v = u \: + at$

$0 = 5 + 6a$

$a = \frac{ - 5}{6} m/ {sec}^{2}$

from newton 2 equation

$s = ut \: + \frac{1}{2} a {t}^{2}$

$where \: s \: = distance$

$s = 5 \times 6 \: + \frac{1}{2} \times \frac{ - 5}{6} \times {6}^{2}$

$s = 30 + ( - 15)$

$s = 15m$