Application of Negative Numbers in Speed Distance Time
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If a body covers a distance x with speed v1 and then it covers another distance x with speed v2, find the average speed for the whole journey.
- 4
- 2
- -6
- -4
- 13
v = 40 km/h. What distance will be covered by the car in 112 hours?
- 60 km
- 50 km
- 80 km
- 70 km
- v=49+9.8t
- v=49−9.8t
- v=9.8−49t
- v=9.8+49t
Write down the units and dimensional formulae of following physical quantities:
Speed
Choose the correct option(s).
- Distance covered at time -2 s is 12 m
- Distance covered at time 4 s is 40 m
- Distance covered at time -2 s is -12 m
- Distance covered at time 0 s is 0 m
A man walked 5 km towards North then 8 km towards South. What is his final position with respect to his initial position?
4 units
-4 units
-3 units
1 unit
- 5 km towards East
- 3 km towards South
- 8 km towards North
- 5 km towards South
A man walked 3 km towards North then 8 km towards South. What is his final position with respect to his initial position?
3 km towards South
8 km towards North
5 km towards South
5 km towards East
- 18
- v=5 ms−1
- v=10 ms−1
- v=15 ms−1
- v=20 ms−1
A man walked 3 km towards North then 8 km towards South. What is his final position with respect to his initial position?
5 km towards East
3 km towards South
8 km towards North
5 km towards South
A train of 150 m in length is going in the north direction at a speed of 10 .A parrot flies at a speed of 5towards the south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to. In the above questions, if the parrot flies at a speed of 5 towards the north. The time taken by the train to leave past the parrot is :
25 s
20 s
30 s
50 s
- 18
Prashanthi travels 30 km towards North then takes about turn and then travels south for 55 km. Taking North direction as positive and south direction as negative, find the prashanthi's distance from the initial starting point.
65 km towards the south.
25 km towards the south.
25 km towards the north.
50 km towards the south.
An elevator descends into a mine shaft at the rate of 15m/min. If the descend starts from 10 m above the ground level, how long will it take to reach -140m.
18 minutes
38 minutes
10 minutes
8 minutes