Question

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

Answer 1

we know that

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

$Speed=\frac{distance}{time}$

in this problem we have

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

substitute in the formula

$Speed=\frac{400}{120}$

$Speed=3.5\frac{m}{sec}$

therefore

the answer Part a) is

the speed is equal to $3.5\frac{m}{sec}$

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

$magnitude=3.5\frac{m}{sec}$

$direction=North\ East\ (NE)$

therefore

the answer Part b) is

the velocity is $3.5\frac{m}{sec}\ North\ East\ (NE)$

Part c)

we know that

the acceleration is equal to the formula

$a=\frac{V2-V1}{t2-t1}$

in this problem we have

$V2=0 \\V1=3.5\frac{m}{sec}$

$t2=15\ sec\\t1=0$

substitute in the formula

$a=\frac{0-3.5}{15-0}$

$a=-\frac{3.5}{15}\frac{m}{sec^{2}}$

$a=-\frac{7}{30}\frac{m}{sec^{2}}$

therefore

the answer Part c) is

the acceleration is $-\frac{7}{30}\frac{m}{sec^{2}}$

This is an example of negative acceleration

Answer 2

Step-by-step explanation:

1. It is given that,

Distance covered by the cyclist, d = 400 m

Time taken, t = 120 s

Speed = distance / time taken

$v=\dfrac{400\ m}{120\ s}$

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 :

$a=\dfrac{v-u}{t}$

$a=\dfrac{0-3.33\ m/s}{15\ s}$  

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

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.