3rd Equation of Motion
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When a car driver travels at the speed of applies brake and brings the car to rest in . then the retardation will be?
- 6 m
- 12 m
- 24 m
- 18 m
- 650 m
- 750 m
- 700 m
- 800 m
- 4 m
- 6 m
- 8 m
- 2 m
- 80/3 ms−1
- 40 ms−1
- 25 ms−1
- 30 ms−1
Derive the kinematic equations of motion.
- Zero
- 8 m/s2
- – 8 m/s2
- 4 m/s2
- 20 ms−2
- 10 ms−2
- 2 ms−2
- 1 ms−2
- 1.5 cm
- 1.0 cm
- 3.0 cm
- 2.0 cm
Write the equations of motion for freely falling body.
A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it ?
A moving train is brought to rest within by applying brakes. Find the initial velocity if the relardation due to brakes is .
- s1=13s2=15s3
- s1=15s2=13s3
- 2s1=13s2=15s3
- s1=16s2=15s3
- 50 m
- 60 m
- 70 m
- 80 m
- √v2−u22
- √v2+u22
- √v2+u24
- √v2−u24
- 1/5
- 2/5
- 3/5
- 4/5
- 1:1
- 1:4
- 1:8
- 1:16
A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and the deceleration are the same, find (a) the true weight of the person and (b) the magnitude of the acceleration. Take g = 9.9 m/s2
- h/9 meters from the ground
- 7h/9 meters from the ground
- 8h/9 meters from the ground
- 17h/18 meters from the ground
- 30 m/s
- 40 m/s
- 50 m/s
- 60 m/s
- 5 sec
- 2.5 sec
- 4 sec
- 10 sec
- 13.75 m
- 18.75 m
- 15.75 m
- 11.75 m
Find the initial velocity of a train, which is stopped in 20 s by applying brakes. The retardation due to brakes is .
- 300 m
- 192 m
- 218 m
- 324 m
- sn
- n
- sn2
- n2s
- t1:t2:t3=1:3:5
- t1:t2:t3=1:2:5
- t1:t2:t3=√3:√2:1
- t1:t2:t3=1:(√2−1):(√3−√2)
- 40 m
- 25 m
- 10 m
- 45 m