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

a) 9.6 m/s²

b) 40.4 m

a) 92.8 m

b) 2.6 m/s

Step-by-step 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 = (v_(f)-v_(o))/(t)`

`a = (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