2. A train appears to move Northwards at a speed of
10√3 ms-1 to a man driving his car eastwards
with speed 10 ms-1. The magnitude of velocity of
the train w.r.t. ground is
(1) 20 m/s
(2) 10/2 m/s
(4) 2012 m/s
(3) 40 m/s
Answers
Answer:
Explanation: velocuity of train wrt car in vector form is 10√3jm/s
And velocity of car in vector form is 10im/s
So by using relative velocity formula
V(train wrt car)= v(train)-v(car)
10√3j=v(train)-10i
V(train)=10i+10√3j
|V(train)|=|10i+10√3j|
|V(train)|=20m/s
20 m/s is the magnitude of the velocity of the train with reference to the ground.
Explanation:
Given that,
The Velocity of the train with reference to man = 10√3 jms^-1
The Velocity of the car with reference to ground = 10 ims^-1
To find,
The magnitude of the velocity of the train with reference to the ground = (i) + (ii)
= |10i+10√3j|
= 20 m/s
As we know,
Velocity train with reference to car = velocity of the train w.r.t. ground - velocity of the car w.r.t ground
10√3 jms^-1 = velocity of the train w.r.t. ground - 10 ims^-1
so,
∵ Velocity of the train w.r.t ground = 10√3 jms^-1 + 10 ims^-1
= |10i+10√3j|
= 20 m/s
Learn more: Magnitude of Velocity
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