Question

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s2 for 7.00 s. (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Answer 1

The final velocity of the racer is 15 m/s. The racer saves 8 seconds by starting to accelerate when he was 300 meters from the finish line. The winner finishes 30.68 meters ahead and 9.27 seconds faster than the other racer.

Step-by-step explanation:

(a) Final velocity:

To find the final velocity, we can use the equation:

v = u + at

Where:

• v is the final velocity
• u is the initial velocity
• a is the acceleration
• t is the time

Substituting the given values:

v = 11.5 + (0.5)(7)

v = 11.5 + 3.5

v = 15 m/s

Therefore, the final velocity is 15 m/s.

(b) Time saved:

To find the time saved, we can use the equation:

s = ut + 0.5at^2

Where:

• s is the distance
• u is the initial velocity
• a is the acceleration
• t is the time

Substituting the given values:

300 = 11.5(7) + 0.5(0.5)(7)^2

Simplifying the equation, we find:

49t^2 - 79t + 300 = 0

(7t - 15)(7t - 20) = 0

Therefore, t = 15 s or t = 20 s. Since the racer accelerates for 7 s, the time saved is 7 - 15 = -8 s.

(c) Distance and time ahead:

The racer's distance and time ahead can be calculated using the equation:

s = ut + 0.5at^2

Substituting the given values:

s = 5 + 11.8(t - 7)

Let's solve for t:

300 = 11.5t + 0.5(0.5)t^2

0.25t^2 + 11.5t - 300 = 0

t = 9.27 s

Substituting t = 9.27 s back into the equation:

s = 5 + 11.8(9.27 - 7)

s = 5 + 25.68

s = 30.68

Therefore, the winner finishes 30.68 m ahead and 9.27 s faster than the other racer.

Answer 2

given,

initial velocity = 11.5 m/s

acceleration = 0.5 m/s²

time = 7 s

a) using equation of motion

v = u + a t

v = 11.5 + 0.5 × 7

v = 15 m/s

b)
`s = u t + (1)/(2)at^2`

`s = 11.5* 7 + (1)/(2)* 0.5* 7^2`

s = 92.75 m

distance traveled by racer with final velocity

d = 300 - 92.75

= 207.25 m

time spend to travel 207.25

`t = (207.25)/(15)`

t = 13.82 s

total time spent to travel 300m

t = 7 + 13.82

t = 20.82 s

time required to travel 300 m with initial velocity

`t_0 =(d)/(v_0)`

`t_0 =(300)/(11.5)`

= 26.09 s

time saved due to acceleration

t = 26.09 - 20.82 = 5.27 s

c) second racer is 5 m apart from the first one

distance traveled by the second racer till first one finishes the race

`d_2 = v_2t`

`d_2 = 11.8* 20.82 = 245.676 m`

distance behind the leader

d = 300 -5 -245.676 = 49.34 m

time behind the winner

`t_f = (d)/(v_2) =(49.34)/(11.8) = 4.18 s`