A particle move with velocity v1 for time t1 and v2 for time t2 along a straight line. the magnitude of its average acceleration is
Answers
Please find the answer below:
Keeping in view that, A particle move with velocity v1 for time t1 and v2 for time t2 along a straight line.
The magnitude of its average acceleration is ( v2 - v1 ) / ( t2 - t1)
This calculated with the formula of acceleration i.e:
Acceleration = ( final velocity - initial velocity )/ (change in time)
Answer:
Explanation:
Time passed before the speed change
speed = d1/t1 ;
65 mi/h = (130 mi)/t1
t1 = 2.0 h.
the time at the lower speed is
t2 = T — t1 = 3.33 h — 2.0 h = 1.33 h.
We discover the separation went at the lower speed from
speed = d2/t2;
55 mi/h = d2/ (1.33 h), which gives d2 = 73 mi.
The complete separation =
D = d1 + d2 = 130 mi + 73 mi = 203 mi.
normal speed = d/t = (203 mi)/(3.33 h) = 61 mi/h.
Note that the normal speed isn’t!
(65 mi/h + 55 mi/h). The two rates were not kept up for
break even with times