Question

a body moving with the uniform accelerate describes 10 metre in 2nd second and 20 metre in 4th second of motion .calculate the distance travelled in 5th second and also initial velocity and acceleration .



pls help​

Answer 1

Answer

• Distance travelled in 5th second is 25 m
• Initial velocity , u = 2.5 m/s
• Acceleration , a = 5 m/s²

Given

• A body moving with the uniform accelerate describes 10 metre in 2nd second and 20 metre in 4th second of motion

To Find

• the distance travelled in 5th second and also initial velocity and acceleration

Concept Used

$\boxed{\begin{minipage}{4cm} \bf Equations\ of\ motion\ :\\\\\rm \bigstar \;\; v=u+at \\\\\rm \bigstar \;\; s=ut+\dfrac{1}{2}at^2\\\\\rm \bigstar \;\; v^2-u^2=2as\\\\\rm \bigstar \;\; S_n=u+\dfrac{a}{2}[2n-1] \end{minipage}}$

Solution

Use 4 th equation of motion .

A/c , " body describes 10 metre in 2nd second  "

$\implies \rm S_2=10\ m\ \& \\\\\rm S_2=u+\dfrac{a}{2}[2(2)-1]\\\\\implies \rm S_2=u+\dfrac{3a}{2}\\\\\implies \rm 10=u+\dfrac{3a}{2}\\\\\implies \rm 20=2u+3a...(1)$

A/c , " body describes 20 metre in 4th second of motion "

$\implies \rm S_4=20\ m\ \& \\\\\rm S_4=u+\dfrac{a}{2}[2(4)-1]\\\\\implies \rm S_4=u+\dfrac{7a}{2}\\\\\implies \rm 20=u+\dfrac{7a}{2}\\\\\implies \rm 40=2u+7a...(2)$

Now , On solving (2) - (1) , we get ,

$\implies \rm 40-20=2u+7a-2u-3a\\\\\implies \rm 20=4a\\\\\implies \rm a=5\ m/s^2$

Now , On sub. a value in (1) , we get ,

$\implies \rm 20=2u+3(5)\\\\\implies \rm 20=2u+15\\\\\implies \rm 2u=5\\\\\implies \rm u=2.5\ m/s$

Let's find distance travelled in 5th second .

$\implies \rm S_5=2.5+\dfrac{5}{2}[2(5)-1]\\\\\implies \rm S_5=2.5+\dfrac{5}{2}[9]\\\\\implies \rm S_5=2.5+2.5(9)\\\\\implies \rm S_5=2.5+22.5\\\\\implies \rm S_5=25\ m$

Answer 2

Answer:

Distance covered in 5th second = 25 m.

Acceleration = 5 m/s^2

Initial velocity = 2.5 m/s.

Explanation:

Let the initial velocity = u

Acceleration = a

FORMULA Used: [ Snth = u+a(2n-1)/2 ]

ATQ, we have

S2nd = 10 and S4th = 20

Using the above formula, we have

10 = u + a(2(2)-1)/2

10 = u + 3a/2.....(1)

20 = u + a(2(4)-1)/2

20 = u + 7a/2....(2)

Subtracting (1) from (2), we get

10 = 4a/2

4a = 20

a = 5 m/s^2

So, Putting a = 5 in eq(1), we get

10 = u +3*5/2

u = 2.5 m/s

Let the distance travelled in 5th seconds be S5th.

So, we have

S5th = 2.5+5(10-1)/2

S5th = 2.5+22.5

S5th = 25 m

So, the distance covered in 5th second = 25 m.