1st Equation of Motion
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A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, the total distance travelled by the car is given by
12(αβα+β)t2
12(α+βαβ)t2
12(α2+β2α)t2
12(α2−β2β)t2
- 14 m/s
- 20 m/s
- 18 m/s
- 60 m/s
- 24 m
- 22 m
- 20 m
- 8 m
(Assume constant acceleration)
- 8 ms−2
- 7 ms−2
- 4 ms−2
- 3 ms−2
A car moving at a speed u is stopped in a certain distance when the brakes produce a deceleration a. If the speed of the car is nu, what must be the deceleration of the car to stop it in the same distance?
√na
na
n2a
n3a
- 54 km
- 30 km
- 24 km
- 47 km
- 20 m/s
- 22 m/s
- 24 m/s
- 18 m/s
- 22 m/s
- 40 m/s
- 30 m/s
- 25 m/s
- 26 m/s
- 28 m/s
- 24 m/s
- 22 m/s
Car A is moving with a speed of 36 km/h on a two-lane road. Two cars B and C, each moving with a speed of 54 km/h in opposite directions on the other lane are approaching car A. At a certain instant when the distance AB = distance AC = 1 km, the driver of car B decides to overtake A before C does. What must be the minimum acceleration of car B so as to avoid an accident?
1 ms−2
2 ms−2
3 ms−2
4 ms−2
- 30
- 60
- 90
- 45
- Its maximum speed is 8 ms−1
- It maximum speed is 6 ms−1
- It travelled a total distance of 24 m
- It travelled a total distance of 18 m
- (α2+β2αβ)t
- (α2−β2αβ)t
- (α+β)tαβ
- αβtα+β
- 54 km
- 30 km
- 24 km
- 47 km
- 26 m/s
- 28 m/s
- 24 m/s
- 22 m/s