Question

A body is moving with speed of 5m/s In 5 sec body comes to rest Find acceleration and distance covered by body

Answer 1

u = 5m/s
v = 0m/s
t = 5 sec

using first equation of motion ,

v = u+at
0=5+a(5)
-5 = 5a
a = -5/5 => -1 m/s². ( negative sign denotes retardation of the a
body )


using second equation of motion,
s = ut+1/2at²
s = 5(5) + 1/2 × -1 × (5)²
s = 25+(-25/2)
s = [25(2) - 25]/2
s = (50-25)/2 => 25/2 m => 12.5 metres

hope this helps

Answer 2

Explanation:

u = 5m / s

v = 0m / s

t=5 sec

using first equation of motion,

v = u + at

0 = 5 + a(5)

- 5 = 5a

a=-5/5=>- m / (s ^ 2) . (negative sign denotes retardation of the a body)

using second equation of motion,

s = ut + 1/2 * a * t ^ 2

s = 5(5) + 1/2 * (- 1) * (5) ^ 2

s = 25 + (- 25/2)

s=(50-25)/2 Rightarrow25/2 m Rightarrow12.5metres

s = [25(2) - 25] / 2