Question

If i move for 10m in 3s, what distance should i cover in the next 2s such that my average speed for the overall journey is 10m/s?

Answer 1

Answer:

50 m

Explanation:

Case 1 : x = 10m, t = 3s => v = 10/3 m/s

Case 2 : x = ? , t = 5s => v = 10 m/s

x = vt = 10×5 = 50 m

Answer 2

Answer:

40 m

Explanation:

As per the provided information in the given question, we have :

• You move 10 m in 3 seconds first.
• You cover next some distance in 2 seconds.
• Required average speed = 10 m/s

We've been asked to calculate the distance covered by you in the next 2s such that your average speed for the overall journey is 10m/s.

Let us suppose the distance covered by you in next 2 seconds be x m. As we know that average speed is given by,

$\longrightarrow \sf{\quad { Speed_{(Avg)} = \dfrac{Total \; distance}{Total \; time} }}$

Here, total distance will be the sum of the distance covered by you in first 3 seconds and next three seconds.

$\longrightarrow \sf{\quad { 10 \; ms^{-1}= \dfrac{(10 + x)\; ms^{-1} }{(3 + 2) \; s} }}$

Preforming addition in the denominator.

$\longrightarrow \sf{\quad { 10 = \dfrac{10 + x }{5} }}$

Now, transposing 5 from from RHS to LHS.

$\longrightarrow \sf{\quad { 10 \times 5 = 10 + x }}$

Performing multiplication.

$\longrightarrow \sf{\quad { 50 = 10 + x }}$

Transposing 10 from RHS to LHS.

$\longrightarrow \sf{\quad { 50 - 10 = x }}$

Performing subtraction.

$\longrightarrow \quad\underline{\boxed { \pmb{\frak{ 40 \; m= x}} }}$

Therefore, you should cover 40 m in next 2 seconds such that your average speed for the overall journey is 10m/s.