Question

a man covers distance in east direction with a speed of 36 km/h
and then turns north and covers the same distance with the speed of 72km/h. Find the average speed and velocity.

Answer 1

Hello!!!!! I can answer this question:

So, initial & final velocities are given:

Initial Velocity, u = 54 km/h = 15 m/sFinal Velocity, v = 63 km/h = 17.5 m/s

change in velocities = v - u = 2.5 m/s

change in time = 1 second

Therefore, acceleration = change in velocities / change in time = 2.5 m/s^2

Since the value of time is not given, Let the time be ‘t’ :

Let Distance covered at time = t be ‘x1’

Let Distance covered at time = (t+1) be ‘x2’

As, x = u * t + 1/2 * a * t^2 ,

Distance covered by car in one second :

= Dist. covered by car in ‘(t+1)’ seconds - Dist. covered by car in ‘t’ seconds

= x2 - x1

=(u * (t+1) + 1/2 * a * (t+1)^2 ) - ( u * t + 1/2 * a * t^2)

= u*t + u + 1/2 * a * ( t^2 + 1 + 2t ) - u * t - 1/2 * a * t^2

= u * t + u + 1/2 * a * t^2 + 1/2 * a + a * t - u * t - 1/2 * a * t^2

= u + 1/2 * a + a * t

Now, u = 15 m/s & a = 2.5 m/s^2

So, u + 1/2 * a + a * t :

= 15 + 1/2 * (2.5) + 2.5*t

= 15 + 1.25 + 2.5*t

=