A car is reduced from 108km/h r to 36km/hr in 4 sec calculate the acceleration
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the answer is 5 m/sec2
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HEY Buddy...!! here is ur answer..
Given that... V1 = 108 km/hr
and V2 = 36 km/hr
Difference in velocity = ∆V = V1 - V2
=> ∆V = 108–36 = 72 km/hr
=> ∆V = 72 km/hr = 72×5/18 = 20 m/sec
And also given that time = 4 sec
Then we know that... Acceleration = ∆V/t
=> Acceleration = 20/4 = 5 m/sec² <<< Ans.
I hope it will be helpful for u..!!
THANK YOU ✌️✌️
Given that... V1 = 108 km/hr
and V2 = 36 km/hr
Difference in velocity = ∆V = V1 - V2
=> ∆V = 108–36 = 72 km/hr
=> ∆V = 72 km/hr = 72×5/18 = 20 m/sec
And also given that time = 4 sec
Then we know that... Acceleration = ∆V/t
=> Acceleration = 20/4 = 5 m/sec² <<< Ans.
I hope it will be helpful for u..!!
THANK YOU ✌️✌️
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