Question

A sprinter runs 100.0 m in 12.2 seconds. If he travels at constant acceleration for the first 50.0 m and then at constant velocity for the final 50.0 m, what was his peak speed?

• 4.10 m/s.

• 8.20 m/s.

• 12.3 m/s.

• 17.5 m/s.

Answer 1

there are two segments

1) constant acceleration

2) constant velocity

for constant acceleration phase,

50 = 1/2 *a * t1²

and

50 = v* t2

also,

v = a * t1 (final velocity after acceleration for 50 m)

so,

50 / t2 = (100 / t1² )* t1

50 / t2 = 100 / t1

100 * t2 = 50 * t1

t1 = 2 * t2

so,

we have t1 + t2 = 12.2

so,

2 * t2 + t2 = 12.2

t2 = 4.0666 seconds

and

t1 = 8.1333 seconds

so,

a = 100 / 8.1333²

a = 1.511 m/s²

so,

peak speed

v = a * t1

v = 1.511 * 8.1333

v = 12.29 m/s

so,

12.3 m/s is the correct answer