Question

A jogger accelerates from rest to 4.55 m/s in 3.64 s. A car accelerates from 15.8 to 25.3 m/s also in 3.64 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 3.64 s?

Answer 1

The answer to your question is:

Step-by-step explanation:

Data

Jogger

vo = 0 m/s

v2 = 4.55 m/s

t = 3.64 s

Car

vo = 15.8 m/s

v2 = 25.3 m/s

t = 3.64

a)

Formula

a =
`(v2 - vo)/(t)`

Jogger

a =
`(4.55 - 0)/(3.64)`

a = 1.25 m/s2

b)

Car

a =
`(25.3 - 15.8)/(3.64)`

a = 2.61 m/s2

c)

Formula d = vot + 1/2 at2

Jogger

d = (0)(3.64) + 1/2(1.25)(3.64)2

d= 8.3 m

Car

d = (15.8)(3.64) + 1/2(2.61)(3.64)2

d = 57.51 + 17.29

d = 74.8 m

Difference = 74.8 - 8.3 = 66.5 m

Answer 2

Listed below

Step-by-step explanation:

Acceleration can be calculated using the following formula:

`A=(V2-V1)/(T)`

Where:

V2= final velocity

V1=initial velocity

T=time.

A) At rest means that the jogger's initial velocity was of 0 m/s. So we get:

`A=(4.55m/s-0m/s)/(3.64)`

`A=1.25 m/s^s`

B)
`A=(25.3m/s-15.8m/s)/(3.64)`

`A=2.60 m/s^2`

C) To calculate the distance traveled we have to use the following formula:

`X= X1+V1. (T2-T1)+(1)/(2).a.(T2-T1)^2`

Where:

X= distance

X1= initial distance (we consider this one as zero as this is not commented on the problem)

V1= initial velocity

T2=final time

T1=initial time

a= acceleration

For the jogger:

`X= 0m/s. (3.64 s)+(1)/(2).1.25m/s^2.(3.64s)^2`

`X= 8.28 m`

For the car:

`X= 15.8m/s. (3.64 s)+(1)/(2).2.60m/s^2.(3.64s)^2`

`X= 74.73 m`

The car travels travels 66.45 m more than the jogger