Question

Samuel is running in a park’s nature trail. At the first checkpoint, he ran 925 m in 10 minutes. At the second checkpoint, he slowed down and only ran 673 m in 13 minutes. What was the average velocity of Samuel’s entire trip (in m/s)? (1 point)

Answer 1

Answer:

The average velocity of Samuel’s entire trip is 1.16 m/s.

Explanation:

Given:

Distance covered at first checkpoint (d₁) = 925 m

Distance covered at second checkpoint (d₂) = 673 m

Time taken for reaching first checkpoint (t₁) = 10 min = 10 × 60 = 600 s [∵1 min = 60 s]

Time taken for reaching second checkpoint (t₂) = 13 min = 13 × 60 = 780 s

Now, the average velocity of Samuel's entire trip is given by the formula:

$\textrm{Average velocity}=\dfrac{\textrm{Total distance covered}}{\textrm{Total time taken}}$

Total distance traveled is equal to the sum of the distances traveled at first and second checkpoints. So,

Total distance covered = $d_1+d_2=925+673=1598\ m$

Total time taken = $t_1+t_2=600+780=1380\ s$

Therefore, the average velocity is given as:

$\textrm{Average velocity}=\dfrac{\textrm{Total distance covered}}{\textrm{Total time taken}}\\\\\textrm{Average velocity}=\frac{1598\ m}{1380\ s}\\\\\textrm{Average velocity}=1.16\ m/s$

Hence, the average velocity of Samuel’s entire trip is 1.16 m/s.