Question

A uniformly accelerated body travels 50 m/s in 5seconds.if it
covers 14m/s during 5th second find out the initial velocity and
acceleration.​

Answer 1

Appropriate Question :

A body moving with uniform acceleration travels 50 m in 5 sec and if it covers 14 m during the 5th second. Find out the initial velocity and acceleration

Answer:

initial velocity = 5 m/s

acceleration = 2 m/s²

Explanation:

Given :

• A uniformly accelerated body travels 50 m in 5 seconds.

• It  covers 14 m during 5th second.

To find :

• the initial velocity

• the acceleration

Solution :

The distance covered in nth second is given by,

 $\boxed{\bf S_n=u+a\bigg( n-\dfrac{1}{2} \bigg)}$

where

u denotes the initial velocity

a denotes the acceleration

• The body covers 14 m during 5th second.

Put n = 5,

 $\sf S_5=u+a \bigg(5-\dfrac{1}{2} \bigg) \\\\ \sf 14=u+a \bigg( \dfrac{10-1}{2} \bigg) \\\\ \sf 14=u+a\bigg( \dfrac{9}{2} \bigg) \\\\ \sf 14=u+\dfrac{9a}{2} \\\\ \sf 14=\dfrac{2u+9a}{2} \\\\ \sf 14(2)=2u+9a \\\\ \sf 2u+9a=28 \longrightarrow [1]$

• It travels 50 m in 5 seconds.

From the second equation of motion,

 $\boxed{\bf S=ut+\dfrac{1}{2}at^2}$

Put t = 5,

 $\sf 50=u(5)+\dfrac{1}{2}(a)(5^2) \\\\ \sf 50=5u+\dfrac{25a}{2} \\\\ \sf 5 \times 10=5 \bigg(u+\dfrac{5a}{2} \bigg) \\\\ \sf 10=u+\dfrac{5a}{2} \\\\ \sf 10=\dfrac{2u+5a}{2} \\\\ \sf 10(2)=2u+5a \\\\ \sf 2u+5a=20 \\\\ \longrightarrow \boxed{\tt 2u=20-5a}$

Substitute the value of 2u in equation [1],

2u + 9a = 28

20 - 5a + 9a = 28

20 + 4a = 28

4(5 + a) = 4(7)

 5 + a = 7

 a = 7 - 5

 a = 2 m/s²

∴ The acceleration of the body is 2 m/s²

Substitute the value of acceleration in equation [1],

2u + 9a = 28

2u + 9(2) = 28

2u + 18 = 28

2u = 28 - 18

2u = 10

u = 10/2

u = 5 m/s

∴ The initial acceleration of the body is 5 m/s

Answer 2

Answer:16

Explanation: