Question

Calculate the acceleration and distance of the body moving with 5 m/s, which comes to rest after traveling for 6 sec?​

Answer 1

Answer:

acceleration= -5/6 m/$s^{2}$

distance = 15 m

Explanation:

Given,

u(initial velocity)=5

t(time)=6

v(final velocity)=0         {because it comes to rest}

Part 1 : Finding acceleration

we have: v=u+at

substituting values we get

0=5+a(6)

a(acceleration)=-5/6

Part 2 : Finding Distance

s(distance)=ut+a$t^{2}$/2

substituting values,

s=5(6)+(-5/6)($6^{2}$)/2

(s)distance =30-15=15

Answer 2

Answer:

Step by step solution

Acceleration =

$v - u \div t$

A body comes to rest, then

$v = 0$

$u = 5$

$t = 6$

$a = 0 - 5 \div 6$

= - 0.83 m/s* or -5/6

Distance = u t +1/2at*

= 5(6) + 1/2(- 5/6)(6)*

= 15m