Question

A car changes its speed from 5m/s to 10m/s and covers a diatance of 1000 m. Find the time taken and acceleration produced????​

Answer 1

GIVEN :-

• Initial velocity (u) = 5m/s
• Final velocity (v) = 10m/s
• Distance covered (s) = 1000m

TO FIND :-

• Time taken (t).
• Acceleration (a).

SOLUTION :-

By 3rd Kinematical equation,

★ v² = u² + 2as

We have ,

• v = 10m/s
• u = 5m/s
• s = 1000m

Putting values,

→ 10² = 5² + 2(a)(1000)

→ 100 = 25 + 2000a

→ 100 - 25 = 2000a

→ 75 = 2000a

→ a = 75/2000

→ a = 0.0375 m/s²

Acceleration of the car is 0.0375m/s².

_________________________

Now ,

By 1st Kinematical equation,

★ v = u + at

We have ,

• v = 10m/s
• u = 5m/s
• a = 0.0375m/s²

Putting values,

→ 10 = 5 + (0.0375)(t)

→ 10 - 5 = 0.0375t

→ 5 = 0.0275t

→ t = 5/0.0375

→ t = 133.33 s

Time taken by the car is 133.33s.

Hence , Acceleration to the car is 0.0375m/s² and Time taken is 133.33s.

MORE TO KNOW :-

• 2nd Kinematical equation

★ s = ut + (1/2)at²

• Kinematical equations are only applicable when acceleration is constant.

Answer 2

Answer:

Answer is = 133.335

Step-by-step explanation:

Given:-

• Initial velocity (u) =5m/s
• final velocity (v) =10m/s
• Distance covered (5)=1000m

To final :-

• Time taken (t)
• Acceleration (a)

solution :-

By 3rd kinematical equation,

V² = u² = 2as

we have,

• v=10m/s
• u=5m/s
• s=1000m

putting values,

:- 10=5²+2(a)(1000)

:- 100 =25+2000a

:- 100-25=2000a

:- 75=2000a

:- a=75/2000

:- a=0.0375mls²

Acceleration of the car is 0.0375m/s².

Now,

by 1st kinematical equation,

v=u+at

we have,

• v=10m/s
• u=5m/s
• a=0.0375

putting values,

:- 10=5+(0.0375)(t)

:- 10-5=0.0375 t

:- 5=0.0275

:- t=5/0.0375

:- t =133.335

Time taken by the car is 133.335.

Hence,Acceleration to the car is 0.0375m/s² and time taken is 133.335 .

more to know:-

• 2nd kinematical equation 5=ut +(1/2)at2
• kinematical equation are only constant.

i hope full help it

thanks for asking