Question

A powerful motorcycle can accelerate from rest to 27.8 m/s (62 mi/h) in only 2.90 s. (a) What is its (constant) acceleration? (b) How far does it travel in that time? -----------------------------------At the end of a race a runner decelerates from a velocity of 9.00 m/s at a rate of 0.400 m/s2. (a) How far does she travel in the next 16.0 s? (b) What is her final velocity?

Answer 1

Answer:

a) 9.6 m/s²

b) 40.4 m

a) 92.8 m

b) 2.6 m/s

Explanation:

a)

$v_{o}$ = initial velocity of motorcycle = 0 m/s

$v_{f}$ = final velocity of motorcycle = 27.8 m/s

t = time taken = 2.90 s

$a$ = acceleration

Acceleration is given as

$a = \frac{v_{f}-v_{o}}{t}$

$a = \frac{27.8-0}{2.90}$

$a$ = 9.6 m/s²

b)

$x$ = distance traveled

Distance traveled is given as

$x = v_{o}t + (0.5)at^{2}$

$x = (0)(2.90) + (0.5)(9.6)(2.90)^{2}$

$x$ = 40.4 m

a)

$v_{o}$ = initial velocity of motorcycle = 9 m/s

$t$ = time taken = 16 s

$a$ = acceleration = - 0.4 m/s²

$x$ = distance traveled

Distance traveled is given as

$x = v_{o}t + (0.5)at^{2}$

$x = (9)(16) + (0.5)(- 0.4)(16)^{2}$

$x$ = 92.8 m

b)

$v_{f}$ = final velocity

final velocity is given as

$v_{f}$ = $v_{o}$ + $a$ t

$v_{f}$ = 9 + (- 0.4) (16)

$v_{f}$ = 2.6 m/s