Question

The velocity-time graph of a ball is given below.What is the acceleration of the ball
(i) During first 10 s?
(ii) Between 10 s and 35 s?
(iii) During 35 s to 40 s?
Explain step by step ​

Answer 1

━━━━━━━━━━━━━━━━━━━━

⏹ Required Answer:

✏ GiveN:

• A velocity time graph showing the velocities at different intervals.

✏ To FinD:

Acceleration in the time interval:

• During first 10 s?
• Between 10 s and 35 s?
• During 35 s to 40 s?

━━━━━━━━━━━━━━━━━━━━

⏹ How to solve?

Question is asking for acceleration and we are provided with velocities at different time intervals. From here we can find initial and final velocity, and the time taken.

✏ Because,

$\large{ \boxed{ \rm accl.n = \dfrac{final \: velocity - initial \: velocity}{time \: taken} }}$

So, by observation we can identify the initial and final velocities, time taken and finding the acceleration by using formula.

━━━━━━━━━━━━━━━━━━━━

⏹ Solution:

(i) Duration of first 10 s

• Initial velocity = 0 m/s
• Final velocity = 10 m/s
• Time interval = 10 s

By using formula,

$\large{ \rm{ \longrightarrow \: a = \frac{v - u}{t} }} \\ \\ \large{ \rm{ \longrightarrow \: a = \frac{10 m {s}^{ - 1} - 0m {s}^{ - 1} }{10 s} }} \\ \\ \large{ \rm{ \longrightarrow \: a = \frac{10 m {s}^{ - 1} }{10s} }} \\ \\ \large{ \rm{ \longrightarrow \: a = \boxed{ \red{ \rm{1m {s}^{ - 2} }}}}}$

(ii) Duration of 10 s and 35 s

• Initial velocity = 10 m/s
• Final velocity = 10 m/s
• Time taken = 35s - 10s = 25s

By using formula,

$\large{ \rm{ \longrightarrow \: a = \frac{10m {s}^{ - 1} - 10m {s}^{ - 1} }{25s} }} \\ \\ \large{ \rm{ \longrightarrow \: a = \frac{0m {s}^{ - 1} }{25s} }} \\ \\ \large{ \rm{ \longrightarrow \: a = \boxed{ \rm{ \red{ 0m {s}^{ - 2} }}}}}$

(iii) Duration of 35s and 40s

• Initial velocity = 10 m/s
• Final velocity = 0 m/s
• Time taken = 40s - 35s = 5s

By using formula,

$\large{ \rm{ \longrightarrow \: a = \frac{0m {s}^{ - 1} - 10m {s}^{ - 1} }{5s} }} \\ \\ \large{ \rm{ \longrightarrow \: a = \frac{ - 10m {s}^{ - 1} }{5s} }} \\ \\ \large{ \rm{ \longrightarrow \: a = \boxed{ \rm{ \red{- 2m {s}^{ - 2} }}}}}$

$\large{ \therefore{ \underline{ \underline{ \purple{ \rm{hence \: solved \: \dag}}}}}}$

━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

Bro you are Aakash student trust your self

By the way answer

1. 1ms^-2
2. 0ms^-2
3. -2ms^-2