Question

The acceleration of a car started at t = 0 , varies with time as shown in figure the distance travelled by the car in the 25s is :-

★ Need a quality Answer.

Note :- Picture in attachment​

Answer 1

EXPLANATION.

The acceleration of a car started at t = 0.

The distance travelled by the car in the 25 seconds is.

As we know that,

Formula of :

Newton 1st equation of motion.

⇒ v = u + at.

Newton 2nd equation of motion.

⇒ S = ut + 1/2 at².

In the first 10 seconds.

Initial velocity = u = 0.

time = t = 10 seconds.

Using newton 2nd equation of motion, we get.

⇒ S = ut + 1/2 at².

⇒ S₁ = 0 x 10 + 1/2 x (5) x (10)².

⇒ S₁ = 1/2 x 5 x 100.

⇒ S₁ = 5 x 50.

⇒ S₁ = 250 ft.

At 10 seconds,

Using newton 1st equation of motion, we get.

⇒ v = u + at.

⇒ v = 0 + 5 x 10.

⇒ v = 50 ft/sec.

As we know that,

Particles moves with a uniform velocity = 50 ft/sec.

From 10 to 20 seconds.

⇒ Δt = t₂ - t₁.

⇒ Δt = 20 - 10.

⇒ Δt = 10 seconds.

Distance covered (S₂) = 50 x 10 = 500 ft.

As we know that,

Between 20 to 25 seconds.

Acceleration is constant = 10 ft/sec².

⇒ Δt = t₂ - t₁.

⇒ Δt = 25 - 20.

⇒ Δt = 5 seconds.

Using newton 2nd equation of motion, we get.

⇒ S = ut + 1/2 at².

⇒ S₃ = 50 x 5 + 1/2 x (-10) x (5)².

⇒ S₃ = 250 + (-5) x (25).

⇒ S₃ = 250 - 125.

⇒ S₃ = 125 ft.

The distance travelled by car in 25 seconds is.

⇒ S = S₁ + S₂ + S₃.

⇒ S = 250 + 500 + 125

⇒ S = 875 ft.

Answer 2

Question :-

⊙ The acceleration of a car started at t = 0 , varies with time as shown in figure the distance travelled by the car in the 25s.

Given :-

⊙ The acceleration of a car started at t = 0 , varies with time as shown in figure the distance travelled by the car in the 25s.

Find Out :-

⊙ What is the distance travelled by the car in the 25s.

Solution :-

☣ In case first 10 second :-

We have ;

• Initial velocity = u = 0.
• time = t = 10 seconds.

As we know that :

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

According to the question,

$\implies$ S = ut + 1/2 at².

$\implies$ S₁ = 0 x 10 + 1/2 x (5) x (10)².

$\implies$ S₁ = 1/2 x 5 x 100.

$\implies$ S₁ = 5 x 50.

$\implies$ ${\small{\bold{\purple{\underline{S_1 = 250\: m}}}}}$

☣ In case first 10 second :-

At 10 seconds,

As we know that :

✯ $\implies$ $\red{ \boxed{\sf{v = u + at}}}$ ✯

$\implies$ v = 0 + 5 x 10.

$\implies$ ${\small{\bold{\purple{\underline{v = 50 m/sec.}}}}}$

☣ In case of total time taken :-

Between 10 to 20 seconds :

$\implies$ Total Time Taken = t₂ - t₁.

$\implies$ Total Time Taken = 20 - 10.

$\implies$ ${\small{\bold{\purple{\underline{Total\: Time\: Taken =\: 10 seconds.}}}}}$

☣ In case of total distance covered :-

$\implies$ Total Distance covered = 50 x 10

$\implies$ ${\small{\bold{\purple{\underline{Total\: Distance\: Covered =\: 500\: m}}}}}$

☣ In case of total time taken :-

Between 20 to 25 seconds.

Acceleration is constant = 10 ft/sec².

$\implies$ Total Time Taken = t₂ - t₁.

$\implies$ Total Time Taken = 25 - 20

$\implies$ ${\small{\bold{\purple{\underline{Total\: Time\: Taken =\: 5\: seconds.}}}}}$

Again,

As we know that :

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

$\implies$ S₃ = 50 x 5 + 1/2 x (-10) x (5)².

$\implies$ S₃ = 250 + (-5) x (25).

$\implies$ S₃ = 250 - 125

$\implies$ ${\small{\bold{\purple{\underline{S_3 = 125\: m}}}}}$

☣ In case of total distance covered :-

The distance travelled by car in 25 seconds is.

$\implies$ Total Distance covered = 250 + 500 + 125

$\implies$ ${\small{\bold{\purple{\underline{Total\: Distance\: Covered =\: 875\: m}}}}}$

Henceforth, the distance travelled by the car is 875 m .