Question

A uniform accelerated motion object covers 65m in 5th second and 105m in 9th . then how much distance will it cover in 20th second ? Also calculate the distance travelled in 20s.

Answer 1

Answer:

• Distance covered by the object in 20 th second = 215 m
• Distance covered by the body in 20 s = 2400 m

Given:

• Distance traveled by the object in 5 th second = 65 m

• Distance traveled by the object in 9 th second = 105 m

Solution:

Distance traveled in n th second is given by,

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

Now,

Distance traveled by the object in 5 th second,

$\sf{S_5= u + \dfrac{a}{2}(2\times 5 -1)}$

$\sf{S_5= u + \dfrac{9a}{2}}$

$\sf{65= u + \dfrac{9a}{2}....(1)}$

Similarly,

Distance traveled by the object in 9 th second,

$\sf{S_9= u + \dfrac{17a}{2}}$

$\sf{105= u + \dfrac{17a}{2}...(2)}$

Subtracting (1) from (2) we get,

$\sf{40= \dfrac{8a}{2}}$

$\sf{40= 4a}$

$\sf{a = 10 m /s^2}$

Now let's substitute the value of a in eq (1),

$\sf{65= u + \dfrac{9\times 10}{2}}$

$\sf{u = 65-45}$

$\sf{u = 20 m/s }$

Now that we have the value of u and a, we can easily find the distance traveled by the object in 20 th second,

$\sf{S_{20}= 20 + \dfrac{10}{2}(2\times 20 -1)}$

$\sf{S_{20}= 20 + 5 \times 39}$

$\boxed{\sf{S_{20}= 215 m }}$

As the object is moving with uniform acceleration we can use the second equation of motion in order to find the distance traveled by the object in 20 s,

$\sf{s= ut + \dfrac{1}{2}at^2}$

$\sf{s= 20 \times 20 + \dfrac{1}{2}\times 10\times 400}$

$\boxed{\sf{s= 2400 \:m }}$