Question

An object initially moving at 12 m/s accelerates uniformly to 33 m/s in 16 s. It continues to move at this speed for 15 s and finally comes to rest with uniform acceleration in 25 s. Calculate: (a) the acceleration in the first 16 s; (b) the total distance moved; (c) the acceleration in the last 10 s; (d) the average speed in the first 30 s.

Answer 1

(a)
the acceleration in the first 16 s
v=u+at
33=12+ a*16
a=21/16 =1.31 m/s²
/
(b)
in first 16 s
S1 = ut+0.5at²
=12*16 +0.5*1.31*16*16=360 m

next 15 seconds
S2 =33*15 =495 m
.
next 25 seconds=S3
v =u +at
0 =33 + a*25
a =- 1.32 m/s²
S3 = 33*25 - 0.5*1.32*25*25
=412.5 m
.
Total distance =S1+S2+S3 =360+495+412.5=1267.5 meters
.
ANS(c)
calculated in part b
a =- 1.32 m/s²

.
ANS(d)
avg speed = total distance/total time =[360+33*14] / 30=822/30= 27.4 m/s

Answer 2

a = 33-12/10 = 21/16 = 1.3125m/sec²

v*v - u*u = 2as

s = 33*33-12*12/1*1.3125 = 360m for 16sec

s = ut = 33*15 = 495m for 15sec

s = u*u/2a = 33*33/2*33/25 = 412.5m

total s = 1267.5m

average speed = [360 + 33*14]/30 = 27.4m/sec

acc in last 10 sec = 33/25 = -1.32m/sec²   negative sign because retardation and given uniform

Answer 3

v = u +at

a)

a1= (33-12)/16

= 1.31m/s²

c)

a2 = (0-33)/25

= -1.32

b)

s = ut + 0.5at²

= 12*16 + 0.5*1.31*16*16 + 33*15 + 33*25 +0.5(-1.32)*25*25

= 1267.8 m

d)

v avg = 27.4 m/s

Answer 4