Question

A body starting from rest acquire a velocity of 10s-1in 5 second calculate (a) the acceleration (b) the distance covered by the body in 5 second

Answer 1

Answer:

15:5:5:5:5:5:5

Explanation:

3:5:4:6:7:8:9

Answer 2

Answer:

a) 2 m/s²

b) 25 m

Explanation:

$\red{\boxed{\rm Solution}}$

As the body is starting from rest,

Therefore,

u = 0 m/s

v = 10 m/s

Time taken to acquire the 10 m/s velocity = 5 s

a) We know that acceleration is the rate of change of velocity per unit time

Mathematical we can write,

$\sf \: a \: = \frac{\Delta v}{t}$

∆v = v - u

Therefore,

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

Applying the above formula,

$\to \: \sf \: a \: = \frac{10 - 0}{5}$

$\to \sf \: a \: = \frac{10}{5} \implies \: a \: = 2 \: m/s^2$

b) We know that intial velocity, final velocity , time taken, acceleration and we need to find the distance covered in 5 seconds

We can use the 2nd equation of motion, i.e s = ut + at²/2 or else we can use the 3rd equation of motion, i.e v² - u² = 2as

• Using 2nd equation of motion:-

We know that the 2nd equation of motion is,

$\sf \: s \: = ut \: + \frac{a {t}^{2} }{2}$

Using the above formula,

$\sf \: s \: = 0 \times 5 + \frac{2 \times 5 \times 5}{2}$

$\sf \: s \: = 25 \: m$

• Using the 3rd equation of motion

We know that the 3rd equation of motion is,

$\sf \: {v}^{2} - {u}^{2} = 2as$

Using the above formula,

$\sf \: {10}^{2} - {0}^{2} = 2 \times 2 \times s$

$\sf \: 100 \: = 4s \implies \: s \: = \: \frac{100}{4} = 25 \: m$