Question

A body starts with an initial velocity of 10 m^-1 per second hand acceleration 5 m/s^-2 find the distance covered by it in 5 seconds

Answer 1

Answer:

Explanation:

INITIAL VELOCITY,U=10 M/S

ACCELERATION= 5 M/S²

TIME TAKEN= 5 SECONDS.

DISTANCE TRAVELLED =?

ACCORDING TO NEWTONS LAWS OF MOTION,

S= UT+0.5 A T²

S=10*5+0.5*5*5*5

S=50+62.5

S=112.5 M .

HOPE IT HELPS.

Answer 2

Solution :

Given :-

▪ Initial velocity of body = 10m/s

▪ Acceleration of body = 5m/s$^2$

To Find :-

▪ Distance covered by body in first five seconds.

Formula :-

▪ As per third equation of kinematics, Formula of distance covered by body in uniform acceleratory motion is given by

$\underline{\boxed{\bf{\pink{s=ut+\dfrac{1}{2}at^2}}}}$

• s denotes distance
• u denotes initial velocity
• t denotes time interval
• a denotes acceleration

Calculation :-

$\implies\bf\:s=ut+\dfrac{1}{2}at^2\\ \\ \implies\sf\:s=(10\times 5)+\dfrac{1}{2}(5\times 5^2)\\ \\ \implies\sf\:s=50+\dfrac{125}{2}\\ \\ \implies\sf\:s=50+62.5\\ \\ \implies\underline{\boxed{\bf{\gray{s=112.5\:m}}}}$