Question

Q 6. Starting from rest a body covers 60 min
10 s then the uniform acceleration of the body
____m/s. *
is
O 1.2
O 2.1
O 3.1
O none​

Answer 1

Given :

▪ Initial velocity = zero (i.e., rest)

▪ Distance covered = 60m

▪ Time interval = 10s

To Find :

▪ Acceleration of body.

Concept :

↗ Since, acceleration has said to be constant throughout the period of motion, we can easily apply second equation of kinematics to solve this question.

✴ Second equation of kinematics :

S = ut + 1/2(at²)

where,

S denotes distance

u denotes initial velocity

a denotes acceleration

t denotes time

Calculation :

→ S = ut + 1/2(at²)

→ 60 = (0×10) + 1/2[a(10)²]

→ 60 = 0 + 1/2(100a)

→ 100a = 120

→ a = 120/100

→ a = 1.2 m/s²

Answer 2

Given:-

● Initial velocity= 0

● Distance covered = 60 m

●Time taken = 10 s

To Find:-

● Uniform acceleration of the body

Concept:-

In a non uniform motion velocity varies with time. It has different values at different instance and at different points of the paths, the change in velocity of the object during any time interval is not zero.

Second equation of kinematics:-

${\tt{\pink{s = ut+\frac{1}{2}at^2}}}$

where,

s = displacement

u= initial velocity

a= acceleration

t = time

Calculation:-

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

$\implies{\bf{60= (0×10) +1/2 [a × (10)^2] }}$

$\implies{\bf{60 = 0 + 1/2( 100a) }}$

$\implies{\bf{100a = 120 }}$

$\implies{\bf{a = 120/100}}$

$\implies{\bf{a = 1.2m/s^2}}$