An electric train moving at 5m/s accelerates to a speed of 8m/s in 20 seconds. Fine the distance travelled in meters during the
2 answers:
You might be interested in
Answer:
The target's velocity is about 1320 m/s in the direction 265.7°.
Explanation:
In order for there to be a collision between missile and target, we must have ...
(target starting position) + (target movement) = (missile movement)
assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...
(10.2 s)(1350 m/s) = 13,770 m
__
We are interested in the target movement, so we can solve for that:
(target movement) = (missile movement) - (target starting position)
In terms of meters, this is ...
(target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°
The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.
As a positive angle, the target's direction is ...
-94.3° +360° = 265.7°
The direction of the target's velocity is 265.7°.
_____
If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.
Using rectangular coordinates, we have ...
13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)
23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)
Then the difference is ...
(12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)
and the (3rd-quadrant) angle is ...
target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°
__
The target's speed is found by dividing the distance it covers by the time it takes.
√(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s
Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.
Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
