A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of \( 7.5 kmh^{-1}\). What is the

**(a)** Magnitude of average velocity, and

**(b)** Average speed of the man over the interval of time

**(i)** 0 to 30 min,

**(ii)** 0 to 50 min,

**(iii)** 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as the magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

Asked by Abhisek | 1 year ago | 167

Distance to the market = 2.5 km = 2500 m

Speed of the man walking to the market= 5 km/h

= 5 x (\( \dfrac{ (5}{18}\)) = 1.388 m/s

Speed of the man walking when he returns back home = 7.5 km/h

= 7.5 x (\( \dfrac{ (5}{18}\)) = 2.08 m/s

**(a)** Magnitude of the average speed is zero since the displacement is zero

**(b)**

**(i)** Time taken to reach the market = \( \dfrac{Distance}{Speed}\)

= \( \dfrac{ 2500}{1.388}\) = 1800 seconds = 30 minutes

So, the average speed over 0 to 30 minutes is 5 km/h or 1.388 m/s

**(ii)** Time taken to reach back home = \( \dfrac{Distance}{Speed}\)

=\( \dfrac{ 2500}{2.08}\) = 1200 seconds = 20 minutes

So, the average speed is

Average Speed over a interval of 50 minutes

= \( \dfrac{Distance\;Covered}{Time\; Taken}\)

= \( \dfrac{ (2500 + 2500)}{3000 }\)

= \( \dfrac{ 5000}{3000}\) = \( \dfrac{5}{3}\) = 1.66 m/s = 6 km/h

**(iii) **Average speed over an interval of 0 – 40 minutes

= \( \dfrac{Distance\;Covered}{Time\; Taken}\)

= \( \dfrac{ (2500 + 1250)}{2400 }\) = 1.5625 seconds = 5.6 km/h

Answered by Pragya Singh | 1 year agoThe velocity-time graph of a particle in one-dimensional motion is shown in the figure.

**(a)** \( x (t_2) = x (t_1) + v (t_1) (t_2 – t_1) +\) (\( \dfrac{1}{2}\)) \( a(t_2 – t_1)^2\)

**(b)** \( v(t_2) = v(t_1) + a(t_2 – t_1)\)

**(c)** Vaverage = \( \dfrac{ x(t_2) – x (t_1)}{(t_2 – t_1)}\)

**(d)** aaverage = \( \dfrac{v(t_2) – v (t_1) }{(t_2 – t_1)}\)

**(e)** \( x (t_2) = x (t_1) + v_{av} (t_2 – t_1) +\) (\( \dfrac{1}{2}\)) \( a_{av} (t_2 – t_1)^2\)

**(f)** \( x(t_2) – x (t_1)\) = Area under the v-t curve bounded by t- axis and the dotted lines.

The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle between

**(a)** t = 0 s to 10 s,

**(b)** t = 2 s to 6 s.

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of \( 15ms^{-1}\) and \(30ms^{-1}\)Verify that the graph shown in given figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = \( 10ms^{-2}\) Give the equations for the linear and curved parts of the plot.

On a long horizontally moving belt figure, a child runs to and fro with a speed 9 km h^{–1} (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^{–1}. For an observer on a stationary platform outside, what is the

**(a)** speed of the child running in the direction of motion of the belt ?.

**(b)** speed of the child running opposite to the direction of motion of the belt?

**(c)** time taken by the child in (a) and (b)?

Which of the answers alter if motion is viewed by one of the parents?

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s^{-1}. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s^{-1} and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?