Question

q2. an object starts from rest and achieves a velocity of 36 km /h in 5 seconds what is the acceleration and displacement during this time ​

Answer 1

Explanation:

first thing first, we have to convert velocity fro km/hr to m/sec

v = 36×1000/3600

v = 10 m/sec

intial velocity, u = 0 m/sec

time = 5 sec

we know that

v = u + at

10 = 0 + a ×5

a = 2 m/s^2

now

s = ut + 1/2at^2

s = 0 + 1/2 × 2 × 25

s = 25 m

Answer 2

Explanation:

Step-by-step explanation:

Distance traveled by it in the last second of motion is 5 m

Given

A ball is thrown vertically upwards with a velocity of 30 m/s

Acceleration due to gravity is 10 m/s²

To Find

distance traveled by it in the last second of motion

Concept Used

As the acceleration due to gravity is constant throughout the motion , so we need to apply equation's of motion ,

⇒ v = u + at

⇒ s = ut + ¹/₂ at²

⇒ v² - u² = 2as

⇒ Sₙ = u + ᵃ/₂ ( 2n-1 )

Solution

Given ,

Initial velocity , u = 30 m/s

Acceleration due to gravity , a = - 10 m/s² [ ∵ thrown against gravity ]

Final velocity , v = 0 m/s [ ∵ stops finally ]

Apply 1st equation of motion ,

⇒ v = u + at

⇒ (0) = (30) + (-10)t

⇒ 10t = 30

⇒ t = 3 s

Now , Distance moved by the particle in nth second is given by ,

$\bf \pink{\bigstar\ \; S_n=u+\dfrac{a}{2}(2n-1)}$★ S

n

=u+

2

a

(2n−1)

u = 30 m/s

n = 3

a = - 10 m/s²

$\begin{gathered}\to \rm S_3=(30)+\dfrac{-10}{2}(2(3)-1)\\\\\to \rm S_3=30-5(5)\\\\\to \rm S_3=30-25\\\\\to \rm S_3=5\ m\end{gathered}$

→S

3

=(30)+ 2−10

(2(3)−1)→S 3

=30−5(5)→S 3

=30−25→S 3

=5 m

Distance traveled by it in the last second of motion is $\textbf{5 m}$