Question

. A body starts from rest and moves with a uniform acceleration of 5m/s2 for 5s and then it moves with a constant velocity for 4s. Later it slows down and comes to rest in 5s. Draw the velocity graph for the motion of the body and answer the following questions: (a) What is the maximum velocity attended by the body? (b) What is the distance travelled during this period of acceleration? (c) What is distance travelled when the body was moving with constant velocity? (d) What is the retardation of the body while slowing down? (e) What is the distance travelled by retarding? (f) What is the total distance travelled?

Answer 1

Given :

✭ First case :

▪ Initial velocity = zero

▪ Acceleration = 5m/s²

▪ Time = 4s

✭ Second case :

▪ Body moves with constant velocity for 4s.

✭ Third case :

▪ Final velocity = zero

▪ Time = 5s

To Find :

A) Max. velocity

B) Distance travelled in first case

C) Distance travelled in second case

D) Retatdation in third case

E) Distance travelled in third case

F) Total distance travelled

Graph :

➳ Refer the attachment.

SoluTion :

✴ First case :

➠ v = u + at

➠ v = 0 + (5×5)

➠ v = 25m/s (Ans. of A)

⇒ v² - u² = 2as

⇒ (25)² - (0)² = 2(5)s

⇒ 625 = 10s

⇒ s = 62.5m (Ans. of B)

✴ Second case :

➜ s' = vt'

➜ s' = 25 × 4

➜ s' = 100m (Ans. of C)

✴ Third case :

➝ v' - v = a't"

➝ 0 - 25 = 5a'

➝ a' = -25/5

➝ a' = -5m/s² (Ans. of D)

➨ v'² - v² = 2(a')s"

➨ (0)² - (25)² = 2(-5)s"

➨ -625 = -10s"

➨ s" = -625/-10

➨ s" = 62.5m (Ans. of E)

➾ Total distace (D) = s + s' + s"

➾ D = 62.5 + 100 + 62.5

➾ D = 225m (Ans of F)

Answer 2

Given :-

First case :-

• Initial velocity = 0
• Acceleration = 5 m/s2
• Time = 4s

Second case :-

• Body moves with constant velocity for 4s

Third case :-

• Final velocity = 0
• Time = 5s

Now here we have to find :-

(a) What is the maximum velocity attended by the body?

(b) What is the distance travelled during this period of acceleration?

(c) What is distance travelled when the body was moving with constant velocity?

(d) What is the retardation of the body while slowing down?

(e) What is the distance travelled by retarding?

(f) What is the total distance travelled?

Solution :-

First case :-

→ v = u + at

→ v = 0 + (5×5)

→ v = 25 m/s

→ v2 - u2 = 2as

→ (25)2 - (0)2 = 2(5)s

→ 625 = 10s

→ s = 62.5 m

Second case :-

→ s' = vt'

→ s' = 25 × 4

→ s' = 100

Third case :-

→ V' - v = a't"

→ 0 - 25 = 5a'

→ a' = -25/5

→ a' = -5m/s2

→ V' - v2 = 2(a')s"

→ (0)2 - (25)2 = 2(-5)s"

→ -625 = -10s"

→ s" = -625/-10

→ s" = 62.5 m

Total Distance :-

→ D = s + s' + s"

→ D = 62.5 + 100 + 62.5

→ D = 225 m