Q27 of 28 Page 55

The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s. What is the average speed of the particle over the intervals in (a) and (b)?

(a) Distance travelled by the particle is equal to the area under s-t graph.


∴ Distance covered in time interval t=0 to t=10 is,


D = 0.5×10×12 m


= 60 m


Average speed = m/s


= = 6 m/s


(b) Since, the particle is undergoing increase in speed in interval t=2 s to t=5 s and decrease in speed in interval t=5 s to t= 6 s. So, we have to deal this interval separately.


Let distance traveled in between 2 s and 5 s be d1 and distance travelled in between 5 s and 6 s be d2. Then,


Total distance covered in t=2 s and t=6 s be,


D = d1 + d2


For distance d1:


For time interval t=0 s to t=5 s,


From 1st equation of motion,


v = u + at


Where,


v = Final velocity = here 12 m/s


u = Initial velocity = here 0 m/s


a = Acceleration/Deceleration = Let a’


t = Time = 5 s


⇒12 = 0 + 5a’


∴ a’ = 2.4 m/s2


Again, from first equation of motion and t = 2 s


v’ = 0 + 2×2.4 = 4.8 m/s


Distance travelled by the particle in t=2 s to t=5s .i.e.,3s


From second equation of motion,


s = ut + 0.5at2


Where,


u = Initial velocity = v’ m/s = 4.8 m/s


a = Acceleration/Deceleration = 2.4 m/s2


s = Distance covered = d1


t = Time = 3 s


⇒ d1 = (4.8×3) + (0.5×2.4×32) = 25.2 m


For distance d2:


For time interval t=5 s to t=10 s,


Final velocity = velocity at 10 s = 0 m/s


Initial velocity = velocity at 5 s = 12 m/s


From 1st equation of motion,


v = u + at


Where,


v = Final velocity = here 0 m/s


u = Initial velocity = here 12 m/s


a = Acceleration/Deceleration = Let a’


t = Time = 5 s


⇒ 0 = 12 + 5a’


∴ a’ = -2.4 m/s2


Distance travelled by the particle in t=5 s to t=6 s..i.e.,1s


From second equation of motion,


s = ut + 0.5at2


Where,


u = Initial velocity = v’ m/s = 12 m/s


a = Acceleration/Deceleration = -2.4 m/s2


s = Distance covered = d1


t = Time = 1 s


⇒ d2 = (12×1) + (0.5×-2.4×12) = 10.8 m


∴ Total distance covered in t=2 s and t=6 s be,


D = d1 + d2 = 25.2 + 10.8 = 36 m


And, Average speed = m/s


= = 9 m/s


More from this chapter

All 28 →
24

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. 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 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?

 25

On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km/h (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. 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?


 26

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in Fig. 3.27 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 = 10 m/s2. Give the equations for the linear and curved parts of the plot.

 28

The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:


Which of the following formulae are correct for describing the motion of the particle over the time-interval t 1 to t 2:


(a) x(t2 ) = x(t1) + v (t1) (t2 – t1) +(1/2) a (t2 – t1)2


(b) v(t2 ) = v(t1) + a (t2 – t1)


(c) vaverage = (x(t2) – x(t1))/(t2 – t1)


(d) aaverage = (v(t2) – v(t1))/(t2 – t1)


(e) x(t2 ) = x(t1) + vaverage (t2 – t1) + (1/2) aaverage (t2 – t1)2


(f) x(t2 ) – x(t1) = area under the v-t curve bounded by the t-axis and the dotted line shown.