calculate the acceleration of the body which was moving with the velocity of 40m/sec and was stopped by applying breaks, also calculate the distance travelled by it in 5 sec
Answers
Given:
- Initial velocity,u = 40 m/s
- Final velocity,v = 0 m/s
- Time taken,t = 5 s
To be calculated:
Calculate the acceleration and distance travelled of the body.
Formula used:
- v = u + at
- s = ut + 1/2at²
Solution:
Case 1 :
From first equation of motion
v = u + at
★ Substituting the values in the above formula,we get
⇒ 0 = 40 + a × 5
⇒ 0 = 40 + 5a
⇒ 5a = -40
⇒ a = -40/5
⇒ a = -8 m/s² [ negative acceleration ]
Thus,the acceleration of given body is -8 m/s².
Case 2:
From Second equation of motion
s = ut + 1/2 at²
s = 40 × 5 + 1/2 × -8 × ( 5 )²
s = 200 + ( -4 ) × 25
s = 200 - 100
s = 100 m
Thus,the distance covered by given body is 100 metres.
✯✯ QUESTION ✯✯
Calculate the acceleration of the body which was moving with the velocity of 40m/sec and was stopped by applying breaks, also Calculate the distance travelled by it in 5 sec ..
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✰✰ ANSWER ✰✰
➥Firstly , We will Find the Acceleration ..
➮Using 1st Equation Of Motion : -
☛Putting Values : -
☛Acceleration of the Body is -8m/s ..
➥Now ,
We will find the distance travelled by the body ..
➥Using 2nd Equation of Motion : -
➥Putting Values : -
☛So , The distance travelled by the body is 100m.