What average force is required to stop an 950-kg car in 8.0 s if the car is traveling at 95 km/h?
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Answer : 3,135 N
Step-by-step explanation :
We know that S.I. unit of speed is m/s but speed given here is in km/hr.
⇒1 km/hr = 5/18 m/s
This is why, 95 km/hr
⇒ 95 × 5/18 m/s = 26.4 m/s
Since finally by applying force we will bring it to velocity of 0 m/s.
⇒v = u + at
⇒0 = 26.4 + a(8)
⇒- 26.4/8 = a
⇒- 3.3 m/s² = a
So, Force = ma
⇒F = 950 kg × ( - 3.3 m/s² )
⇒F = - 3,135 N
Here negative sign shows that force will be applied against direction of motion.
Step-by-step explanation :
We know that S.I. unit of speed is m/s but speed given here is in km/hr.
⇒1 km/hr = 5/18 m/s
This is why, 95 km/hr
⇒ 95 × 5/18 m/s = 26.4 m/s
Since finally by applying force we will bring it to velocity of 0 m/s.
⇒v = u + at
⇒0 = 26.4 + a(8)
⇒- 26.4/8 = a
⇒- 3.3 m/s² = a
So, Force = ma
⇒F = 950 kg × ( - 3.3 m/s² )
⇒F = - 3,135 N
Here negative sign shows that force will be applied against direction of motion.
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