Question

A body moves with a constant velocity of 5m/s for 10s and then decelerates at the rate 1m/s^2

Answer 1

l l Correct Question l l

A body moves with a constant velocity of 5m/s for 10s and then decelerates at the rate 1m/s. Calculate the distance covered by moving body ?

Given:

• Initial velocity,u = 5 m/s

• Time taken, t = 10 s

• Acceleration, a = 1 m/s²

To be calculated:

Calculate the distance covered by given body?

Formula used:

s = ut + 1/2 at²

Solution:

According to the second equation of motion,

s = ut + 1/2 at²

☆ Substituting the value in the above formula, we get

s = 5 × 10 + 1/2 × 1 × ( 10 )²

s = 50 + 1/2 × 100

s = 50 + 50

s = 100 m

Thus, the total distance covered by given body is 100 m.

Answer 2

$\huge{\underline{\purple{\tt{Appropriate \ Question :}}}}$

$\longrightarrow$ A body moves with a constant velocity of 5m/s for 10s. and then decelerates at the rate 1m/s². Find the total distance covered by the body ?

$\rule{200}1$

$\huge{\underline{\pink{\tt{Given,}}}}$

• A body moves with a Constant Velocity (Initial Velocity) = 5 m\s
• Time Taken = 10 seconds
• Acceleration = 1 m\s²

$\huge{\underline{\pink{\tt{To \ Find,}}}}$

• The Total Distance Covered by the Body

$\huge{\underline{\pink{\tt{Solution :}}}}$

$\implies \underline{\sf{According\:to\:the\:Second\:Equation\:of\:Motion :}}$

$\mapsto \boxed{\sf{\red{s = ut + \dfrac{1}{2} at^{2}}}}$

Now Put the Value of Given data in this Equation :

$\longrightarrow \sf{5 \times 10 + \dfrac{1}{2} \times 1 \times (10)^{2}}$

$\longrightarrow \sf{s = 50 + \dfrac{1}{2} \times 100}$

$\longrightarrow \sf{s = 50 + 50}$

$\longrightarrow \boxed{\sf{s = 100 m}}$

Therefore,

$\longrightarrow$ The Total Distance Covered by the Body = 100 m

$\rule{200}2$