Bittu starts his car from rest from his house with acceleration of 6 m/s^2 .If he took 7 second to reach his destiny .Calculate the distance of destiny from his house .
Question
Bittu starts his car from rest from his house with an acceleration of 6 m/s². If he took 7 seconds to reach his destiny. Calculate the distance of destiny from his house.
___________________________
Answer
Initial Velocity (u) = 0 m/s
Acceleration (a) = 6 m/s²
Time (t) = 7 sec
Distance (s) = ?
To find the distance covered, we'll apply the 2nd Equation of Motion.
2nd Equation of Motion → s = ut + \dfrac{1}{2}at^{2}s=ut+
2
1
at
2
Substitute the values of variables from the equation and find the value of 's'.
\implies s = 0(7) + \dfrac{1}{2}(6)(7)^{2}⟹s=0(7)+
2
1
(6)(7)
2
\implies s = 0 + 3 \times 49⟹s=0+3×49
\implies s = 147 \ m⟹s=147 m
∴ The distance of destiny from Bittu's house was 147 m.
___________________________
Additional Information
Acceleration is the rate of change of velocity in a given time period.
We have 3 Equations of Motion. They are -:
1st Equation of Motion -: v = u + at
2nd Equation of Motion -: s = ut + 1/2 at²
3rd Equation of Motion -: v² - u² = 2as
Here,
'u' stands for Initial Velocity
'v' stands for Final Velocity
't' stands for Time
'a' stands for Acceleration
's' stands for Distance
___________________________
Answers
Answer:
Bittu starts his car from rest from his house with acceleration of 6 m/s^2 .If he took 7 second to reach his destiny .Calculate the distance of destiny from his house .
Step-by-step explanation:
Answer
Initial Velocity (u) = 0 m/s
Acceleration (a) = 6 m/s²
Time (t) = 7 sec
Distance (s) = ?
To find the distance covered, we'll apply the 2nd Equation of Motion.
2nd Equation of Motion → s = ut + \dfrac{1}{2}at^{2}s=ut+
2
1
at
2
Substitute the values of variables from the equation and find the value of 's'.
\implies s = 0(7) + \dfrac{1}{2}(6)(7)^{2}⟹s=0(7)+
2
1
(6)(7)
2
\implies s = 0 + 3 \times 49⟹s=0+3×49
\implies s = 147 \ m⟹s=147 m
∴ The distance of destiny from Bittu's house was 147 m.
___________________________
Additional Information
Acceleration is the rate of change of velocity in a given time period.
We have 3 Equations of Motion. They are -:
1st Equation of Motion -: v = u + at
2nd Equation of Motion -: s = ut + 1/2 at²
3rd Equation of Motion -: v² - u² = 2as
Here,
'u' stands for Initial Velocity
'v' stands for Final Velocity
't' stands for Time
'a' stands for Acceleration
's' stands for Distance
___________________________
Answer:
Step-by-step explanation:
Initial Velocity (u) = 0 m/s
Acceleration (a) = 6 m/s²
Time (t) = 7 sec
Distance (s) = ?
To find the distance covered, we'll apply the 2nd Equation of Motion.
2nd Equation of Motion → s = ut + \dfrac{1}{2}at^{2}s=ut+
2
1
at
2
Substitute the values of variables from the equation and find the value of 's'.
\implies s = 0(7) + \dfrac{1}{2}(6)(7)^{2}⟹s=0(7)+
2
1
(6)(7)
2
\implies s = 0 + 3 \times 49⟹s=0+3×49
\implies s = 147 \ m⟹s=147 m
∴ The distance of destiny from Bittu's house was 147 m.
___________________________
Additional Information
Acceleration is the rate of change of velocity in a given time period.
We have 3 Equations of Motion. They are -:
1st Equation of Motion -: v = u + at
2nd Equation of Motion -: s = ut + 1/2 at²
3rd Equation of Motion -: v² - u² = 2as
Here,
'u' stands for Initial Velocity
'v' stands for Final Velocity
't' stands for Time
'a' stands for Acceleration
's' stands for Distance