Question

9. A body moves along a straight line with position given by x=8t−2t2 ,where x is in meters and t is in seconds. Find the average velocity and average speed of the body in the intervals: (a) from ti =0 to tf =2s, and (b) from ti =0 to tf =5s.

Answer 1

Answer:

for a ans is 4m/s

for b ans is -2m/s (velocity acting opposite direction of motion)

and here average speed and velocities are same

Answer 2

Given :  A body moves along a straight line with position given by

x=8t−2t²,  where x is in meters and t is in seconds.

To Find :   the average velocity and average speed of the body in the intervals:

(a) from ti =0 to tf =2s, and (b) from ti =0 to tf =5s.

Solution:

First find average velocity  =Displacement / time

Displacement = Final position 0- Initial position

x=8t−2t²

(a) from ti =0 to tf =2s

x (0) = 0

x(2) = 8(2) - 2(2)² = 8

Average velocity = 8/(2 - 0)  = 4 m/s

(b) from ti =0 to tf =5s.

x (0) = 0

x(5) = 8(5) - 2(5)² = -10

Average velocity = =10/(5 - 0)  = -2 m/s

Now find average speed =  Distance/ time

x=8t−2t²

dx/dt  = 8 - 4t  

t < 2  velocity is +ve

t > 2 Velocity is - ve

(a) from ti =0 to tf =2s   , average speed will be be same  as average velocity

as velocity direction is not changing

Hence = 4 m/s

(b) from ti =0 to tf =5s.

need to break  in 2 0 to 2 secs and  2  to 5 secs

Distance in 0 to 2 secs  is   8 m

Distance from 2 to 5 secs   = | -10 - 8 | = 18 m

Hence Distance = 8 + 18 = 26 m

Average speed = 26/5  = 5.2  m/s

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