Determine the dimensions of the derivative dx/dt = 3At^2 + B. (Use the following as necessary: L and T, where L is the unit of l
1 answer:
Answer:
LT⁻¹
Explanation:
Assuming the given expression is
x = A t³ + B t.........(1)
x is the distance
we have to calculate dimension of dx/dt
from expression (1)
x = A t³
A = LT⁻³
x = B t
B = LT⁻¹
now,
dx/dt = 3At^2 + B
from rule of dimension
dimension of dx/dt is equal to dimension of At^2
dx/dt = A t²
dx/dt = LT⁻³ x T²
dx/dt = LT⁻¹
hence, dimension of dx/dt is equal to LT⁻¹
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Answer:
The trains will collide at a distance 1660 m from the station
Explanation:
Let the train traveling due north with a constant speed of 100 m/s be Train A.
Let the train traveling due south with a constant speed of 136 m/s be Train B.
From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).
Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.
∴ If train A has covered a distance m by the time of collision, then train B would have traveled
m.
Also,
At the position where the trains will collide, the two trains must have traveled for equal time, t.
That is, At the point of collision,
is the time spent by train A
is the time spent by train B
From,
Since the time spent by the two trains is equal,
Then,
m
m
m/s
m/s
Hence,
≅ 1,221 m
This is the distance covered by train A by the time of collision.
Hence, Train B would have covered (2881 - 1221)m = 1660 m
Train B would have covered 1660 m by the time of collision
Since it is train B that leaves a station,
∴ The trains will collide at a distance 1660 m from the station.