Question

A body starts with an initial velocity of 10 m s-^1 and accelerates 5 m s-^2. Find the distance covered in 5secs.

Answer 1

Answer:

Given,

u=10m/s

a=5m/s2

t=5sec

From 2nd equation of motion,

s=ut= 1/2at2

s= 10*5+  1/2*5*5*5

s=50+62.5

=112.5

Thank you

Answer 2

$\boxed{\rm{\blue{Find \longrightarrow }}}$

Distance Covered?

$\boxed{\rm{\orange{Given \longrightarrow }}}$

• Initial velocity [u] = 10m/s
• Acceleration [a] = 5m/s²
• Time [t] = 5 sec

$\boxed{\rm{\pink{Formula\ used \longrightarrow }}}$

using Second equation of motion,

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

• where s= distance
• u = initial velocity
• t= time taken
• a= acceleration

$\boxed{\rm{\red{SoLution \longrightarrow }}}$

Put value in the formula we get,

$\sf{ s= 10\times 5 + \dfrac{1}{2} \times 5 \times 5\times 5}$

$\sf{ s= 50 + 62.5}$

$\sf{s= 112.5 m}$

$\underline{\bf{\bigstar \therefore Distance\ covered = 112.5m \bigstar }}$

________________________________