Question

Question:-) A Scooter is traveling at a speed of 36km/hr
after one second, it's velocity becomes 45km/hr. calculate the acceleration experienced by the scooter and the distance covered by it in one second?

don't post irrelevant answer ...​

Answer 1

Answer:

Given:

Initial velocity of scooter,u= 36 km/h

Final velocity of scooter,v= 45 km/h

Time taken by scooter

To Find:

The acceleration experienced by the scooter

Solution:

We know that,

According to first equation of motion for constant acceleration,

\purple{\boxed{\bf{v=u+at}}}

v=u+at

where,

v is final velocity

u is initial velocity

a is acceleration

t is time taken

━━━━━━━━━━━━━━━━━━━━

It is given that,

\longrightarrow\sf{u=36\ km/h=36\times\dfrac{5}{18}\ m/s=10\ m/s}⟶u=36 km/h=36×

18

5

m/s=10 m/s

\longrightarrow\sf{v=45\ km/h=45\times\dfrac{5}{18}\ m/s=\dfrac{25}{2} \ m/s}⟶v=45 km/h=45×

18

5

m/s=

2

25

m/s

━━━━━━━━━━━━━━━━━━━━

Let the acceleration of scooter be 'a'

So, on applying first equation of motion on scooter, we get

\longrightarrow\sf{v=u+at}⟶v=u+at

\longrightarrow\sf{\dfrac{25}{2} =10+a(1)}⟶

2

25

=10+a(1)

\longrightarrow\sf{\dfrac{25}{2}-10=a}⟶

2

25

−10=a

\longrightarrow\sf{\dfrac{25-20}{2}=a}⟶

2

25−20

=a

\longrightarrow\sf{\dfrac{5}{2}=a}⟶

2

5

=a

\longrightarrow\sf{a=\dfrac{5}{2}}⟶a=

2

5

\longrightarrow\sf\green{a=2.5\ m/s^{2}}⟶a=2.5 m/s

2

Hence, the acceleration experienced by the scooter is 2.5 m/s².

Answer 2

Explanation:

The answer is 2.5m/s square