Answer:
<em>1,378.9ms²</em>
Explanation:
Given the following
Distance S = 70.6m
Time t = 0.32secs
Initial velocity = 0m/s
Required
Acceleration
Using the equation of motion
S = ut+1/2at²
Substitute
70.6 = 0+1/2a(0.32)²
70.6 = 0.0512a
a = 70.6/0.0512
a = 1,378.9
<em>Hence the acceleration is 1,378.9ms²</em>
Answer:

Explanation:
First of all, let's convert from nanometres to metres, keeping in mind that

So we have:

Now we can convert from metres to centimetres, keeping in mind that

So, we find:

Answer:
Part a)

Part b)
North of East
Explanation:
Speed of train towards East = 60 km/h
displacement towards East is given as

now it turns towards 50 degree East of North
so its distance is given as


then finally it moves towards west for 50 min

Now the total displacement of the train is given as



now total time duration of the motion is given as


now average velocity is given as


Part a)
magnitude of the average velocity is given as



Part b)
Direction of the velocity is given as


North of East