A 100 m sprinter increases her speed from rest uniformly at the rate of 1 m/s2 upto 40 m and covers the remaining distance with a uniform speed. The sprinter covers the first half of the run in t1s and second half in t2s, then
For the first 40 m:
Using the third equation of motion: v2 – u2 = 2as
where: s = distance covered = 40 m
v = final velocity = ? m/s
u = initial velocity = 0 m/s
a = acceleration = 1 m/s2
v2 – u2 = 2as
v2 – 0 = 2 × 1 × 40
v2 = 80
v = 8.94 m/s
Using the first equation of motion: v = u + at
where: v = final velocity = 8.94 m/s
u = initial velocity = 0 m/s
a = acceleration = 1 m/s2
t1 = time = ?s
v = u + at1
8.94 = 0 + 1 × t1
t1 = 8.94s
Here its given the speed is uniform not velocity, therefore
acceleration will be 1 m/s2 and speed will be 8.94 m/s everywhere.
Using the third equation of motion: v2 – u2 = 2as
where: s = distance covered = 60 m
v = final velocity = ?? m/s
u = initial velocity = 8.94 m/s
a = acceleration = 1 m/s2
v2 – u2 = 2as
v2 – (8.94)2 = 2 × 1 × 60
v2- 80 = 120
v2 = 120 + 80 = 200
v = 14.14 m/s
Using the first equation of motion: v = u + at
where: v = final velocity = 14.14 m/s
u = initial velocity = 8.94 m/s
a = acceleration = 1 m/s2
t2 = time = ?s
v = u + at2
14.14 = 8.94 + 1 × t2
t2 = 14.14 – 8.94
t2 = 5.2 secs