Question

A car moves with velocity 10 m/s towards East in
5 s, its velocity changes to 10 m/s towards North.
Then the change in velocity of car is
(1) 10V2 m/s, North-East
(2) 10/2 m/s, North-West
(3) 10V2 m/s, South-East
(4) 10/2 m/s, South-West​

Answer 1

Answer:

THE ANSWER TO THIS QUESTION IS OPTION 2)

Explanation:

GIVEN:

• Initial velocity of the car is= 10 m/s East = 10 m/s i = v₁

• Final velocity of the car is=  10 m/s North = 10 m/s j = v₂

Now to find the change in velocity of the car we use vector change i.e

CHANGE = v₂-v₁

                =10 j - 10 i

MAGNITUDE OF A VECTOR OF THE FORM

x i + y j = $\sqrt{x^{2} +y^{2} }$

Taking the magnitude of the vector change = $\sqrt{10^{2}+10^{2}  }$=10$\sqrt{2}$

The direction of the change we can say by looking at initial and final states as its tendency is inclined towards moving in the north and west direction.

The time mentioned in the problem is not required to solve it as we are using vector methods to solve the problem.

Normally we could also use scalars to use and herein we can find by using the time.

Hence we can find the answer to be option (4).

Answer 2

Car  Moves in North West Direction

Explanation:

• Average acceleration is defined as- (change in velocity/time elapsed)

• Here,change in velocity = $\sqrt 10^2 + 10^2\ or\ 10\sqrt 2\ ms^-^1$

So,average acceleration is $\frac {10\sqrt 2}{10}\ i.e.\ \sqrt 2\ ms^-^2$

• As,both the velocity vectors are equal in magnitude,their resultant will be at angle of 45° w.r.t each other. And that will be the direction of the average acceleration as well,i.e along north west

Note: **Change in acceleration means,final velocity vector(v) - initial velocity vector (v')

= v + (-v')

• As v' vector is located towards east, so −v' will be located towards west, so a resultant of two vectors of equal magnitude one directed along the north and one along west will be along northwest