Question

A cyclist travels at distance of 400 meters in 120 seconds towards school, calculate his speed. (Show your work)

Calculate his velocity if the direction is North East (NE)
 

The cyclist in practice question 2, comes to a stop position within 15 seconds. Calculate his acceleration. (Show your work) Is this an example of positive or negative acceleration?

Answer 1

Explanation:

1. It is given that,

Distance covered by the cyclist, d = 400 m

Time taken, t = 120 s

Speed = distance / time taken

`v=(400\ m)/(120\ s)`

v = 3.33 m/s

So, the speed of the cyclist is 3.33 m/s.

2. The cyclist comes to a stop position within 15 seconds. Its acceleration is given by :

`a=(v-u)/(t)`

`a=(0-3.33\ m/s)/(15\ s)`

`a=-0.22\ m/s^2`

So, the acceleration of the cyclist is
`-0.22\ m/s^2`. It is an example of negative acceleration as the cyclist is decelerating. Hence, this is the required solution.

Answer 2

we know that

The scalar magnitude of the velocity vector is the speed. The speed is equal to

`Speed=(distance)/(time)`

in this problem we have

`distance=400\ m \\time=120\ sec`

substitute in the formula

`Speed=(400)/(120)`

`Speed=3.5(m)/(sec)`

therefore

the answer Part a) is

the speed is equal to
`3.5(m)/(sec)`

Part b) Find the velocity

we know that

Velocity is a vector quantity; both magnitude and direction are needed to define it

in this problem we have

the magnitude is equal to the speed

`magnitude=3.5(m)/(sec)`

`direction=North\ East\ (NE)`

therefore

the answer Part b) is

the velocity is
`3.5(m)/(sec)\ North\ East\ (NE)`

Part c)

we know that

the acceleration is equal to the formula

`a=(V2-V1)/(t2-t1)`

in this problem we have

`V2=0 \\V1=3.5(m)/(sec)`

`t2=15\ sec\\t1=0`

substitute in the formula

`a=(0-3.5)/(15-0)`

`a=-(3.5)/(15)(m)/(sec^(2))`

`a=-(7)/(30)(m)/(sec^(2))`

therefore

the answer Part c) is

the acceleration is
`-(7)/(30)(m)/(sec^(2))`

This is an example of negative acceleration