Question

A ball starts from rest and rolls down a hill with uniform acceleration, traveling 160 m during the second 5.1 s of its motion.

How far did it roll during the first 5.1 s of motion?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Consider the initial velocity of the ball = vo,
velocity after first 5.1s = v1,
velocity after second 5.1s = v2

vo = 0
v1 = vo + 5.1 a = 0 + 5.1 a
v1 = 5.1 a------------(i)
v2 = v1 + 5.1 a---------(ii)
Substituting the value of v1 from (i) into (ii) -
v2 = 5.1 a + 5.1 a
v2 = 10.2 a------------(iii)

Distance traveled in the second 5.1s = [(v1+v2)/2] * 5.1
Therefore [(v1+v2)/2] * 5.1 = 160
v1 + v2 = 62.7 ------------(iv)
Substituting the values of v1 and v2 from (i) and (iii) into (iv),
5.1a + 10.2a = 62.7
15.3a = 62.7
=> a = 62.7/15.3 = 4.10 m/s^2
So, distance traveled in the first 5.1 s
= vo*5.1 + (1/2) * a* 5.1^2
= 0 + (1/2) * 4.10 * 5.1^2
= 53.32 m

Answer in two significant digits = 53.0 m.