Question

Find the distance covered by the body with an initial velocity of 10 m/s having an acceleration of 10m/s^2 during the 4 th sec of its motion.​

Answer 1

Answer:

sorry I don't know answer

Answer 2

Solution:

Understanding the question: We are asked to find out the distance covered by the body with an initial velocity of 10 mps having an acceleration of 10 mps sq. during the 4th second of its motion.

Provided that:

• Second = 4th
• Initial velocity = 10 mps
• Acceleration = 10 mps sq.

To determine:

• Distance travelled

Solution:

• Distance travelled = 45 m

Using formula:

• ${\small{\underline{\boxed{\pmb{\sf{S_n \: = u + \dfrac{1}{2} a(2n-1)}}}}}}$

Required solution:

$:\implies \tt S_n \: = u + \dfrac{1}{2} a(2n-1) \\ \\ :\implies \tt S_n \: = 10 + \dfrac{1}{2} \times 10[2(4)-1] \\ \\ :\implies \tt S_n \: = 10 + \dfrac{1}{2} \times 10(8-1) \\ \\ :\implies \tt S_n \: = 10 + \dfrac{1}{2} \times 10(7) \\ \\ :\implies \tt S_n \: = 10 + \dfrac{1}{2} \times 70 \\ \\ :\implies \tt S_n \: = 10 + 1 \times 35 \\ \\ :\implies \tt S_n \: = 10 + 35 \\ \\ :\implies \tt S_n \: = 45 \: m$

• Therefore, distance travelled = 45 metres.