Question

A body starting from rest and moving with uniform acceleration acquires a belocity of 16m/s in 4 second, find its acceleration and the distance covered by it in 4 sec.

Answer 1

Answer:

acceleration, a = 4 m/s^2

distance covered, s = 32 m

Step explanation in attachment.

Answer 2

Given,

A body starts from rest, moving with uniform acceleration,

In time = 4 sec, acquires a velocity = 16 m/s.

To find,

Acceleration and distance covered in 4 sec.

Solution,

The relation between velocity, acceleration, and time is given as,

$v= u+at$     ...(1)

and that among distance (or displacement), initial velocity, and acceleration is,

$s= ut+\frac{1}{2} at^2$     ...(2)

Where,

u = initial velocity,

v = final velocity,

a = acceleration,

t = time,

s = distance.

Using the above relations, we can find the acceleration first, as follows.

Here, since body is starting from rest,

u = 0 m/s,

v = 16 m/s (given),

t = 4 s (given).

From relation (1),

$a=\frac{v-u}{t}$

⇒ $a=\frac{16-0}{4}=\frac{16}{4}$

⇒ a = 4 m/s².

Now, to determine the distance, we can use the relation from equation (2), and the above-determined value of a = 4 m/s². So,

$s=(0) \times (4)+\frac{1}{2}\times 4\times (4)^2$

⇒ $s=2\times 16$

⇒ s = 32 m.

Therefore, the acceleration of the given body will be 4 m/s², and the distance covered by the body in 4 sec will be 32 m.