Question

solve the marked ones​

Answer 1

Answer:

v1=10/2=5 u=25/5

=》a=v-u/t

=0

therefore velocity =5

d=7×5=35 and

v=25/5=5

u=34/8=4.25

a=.75/3

s=ut+1/2at2=44.5m

Explanation:

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Answer 2

Question :

1. A body covers 10 m in the 2nd second and 25 m in 5th second .If the motion is uniformly accelerated, how far will it go in the seventh second?
2. An object is moving with uniform acceleration. Its velocity after 5s is 25 m/s and after 8s is 34m/s. Find the distance travelled by the object in 12th second.

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion.

$\bf\:v=u+at$

$\bf\:S=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2aS$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

Solution :

1) We have ,

• Distance covered in 2nd second $\sf\:S_2=10m$
• Distance covered in 5th second $\sf\:S_5=25m$

We have to find ,Distance travelled by the body in 7th second .

Let the initial velocity of the body be u and Acceleration be a .

We know that

$\bf\green{S_{nth}=u+\dfrac{a}{2}(2n-1)}$

Thus ,

$\sf\:S_{2nd}=u+\dfrac{a}{2}[2(2)-1]$

$\sf\implies\:S_{2nd}=u+\dfrac{a}{2}(4-)$

$\sf\implies\:S_{2nd}=u+\dfrac{a}{2}(3)$

$\sf\implies10=u+\dfrac{3a}{2}\:\:equation(1)$

And

$\sf\:S_{5th}=u+\dfrac{a}{2}[2(5)-1]$

$\sf\implies\:S_{5th}=u+\dfrac{a}{2}[10-1]$

$\sf\implies\:S_{5th}=u+\dfrac{a}{2}(9)$

$\sf\implies25=u+\dfrac{9a}{2}\:\:equation(2)$

Now , Subtract equation (1) from (2) ,Then

$\sf\:25-10=\dfrac{9a}{2}-\dfrac{3a}{2}$

$\sf\implies15=\dfrac{6a}{2}$

$\sf\implies\:a=5ms^{-2}$

Put the value of a = 5 m/s² in equation (2)

then , u = 5/2 m/s

Then ,

$\sf\:S_{7th}=u+\dfrac{a}{2}[2(7)-1]$

$\sf\implies\:S_{7th}=\dfrac{5}{2}+\dfrac{a}{2}(13)$

$\sf\implies\:S_{7th}=\dfrac{5}{2}(1+13)$

$\sf\implies\:S_{7th}=35m$

Therefore , the body will go 35 m in 7th second .

2) We have to find the distance travelled by the object in 12th second.

Let the initial velocity of the object be u and Acceleration be a .

According to the Question :

• Velocity of object after 5sec is 25 m/s

• and after 8 sec , it is 34 m/s

By Equation of Motion

25 = u+5a equation(1)

and 34=u+8a equation(2)

Now , Subtract equation (1) from (2) , then

$\sf\:34-25=8a-5a$

$\sf\implies\:a=3ms^{-2}$

Put a=3 in equation (1)

Then , u = 10 m/s

Distance travelled in 12th second

$\sf\:S_{12th}=u+\dfrac{a}{2}[2(12)-1]$

$\sf\implies\:S_{12th}=10+\dfrac{3}{2}(23)$

$\sf\implies\:S_{12th}=10+\dfrac{69}{2}$

$\sf\implies\:S_{12th}=\dfrac{89}{2}$

$\sf\implies\:S_{12th}=44.5m$

Therefore, the distance travelled by the object in 12th second is 44.5 m.