Question

A van moving with uniform acceleration and changes it's velocity from 20m/s to 60m/s in 5s.Find out the displacement in this time ?​

Answer 1

Answer

• The distance covered will be 200 m

Explanation

Given

• Initial Velocity of the van is 20 m/s
• Final Velocity of the van is 60 m/s
• Time taken is 5 sec

To Find

• The distance covered during this time

Solution

• We shall first find the acceleration of the body by using the first equation of motion. Then with the help of the second or the third equation of motion we may find the distance covered

✭ Acceleration of the van

→ v = u+at

→ 60 = 20 + a×5

→ 60-20 = 5a

→ 40 = 5a

→ 40/5 = a

→ Acceleration = 8 m/s²

✭ Distance Covered by the van

● Here we may use two methods, one is to use the second equation of motion and the other is to use the third equation of motion and we'll get the same answer from both the cases.

Method 1

→ s = ut+½at²

→ s = 20 × 5 + ½ × 8 × 5²

→ s = 100 + 4×25

→ s = 100+100

→ Distance = 200 m

Method 2

→ v²-u² = 2as

→ 60²-20² = 2×8×s

→ 3600-400 = 16s

→ 3200 = 16s

→ 3200/16 = s

→ Distance = 200 m

Answer 2

Answer:

$\huge{ \underline{ \large{ \rm{ \pink{Given:}}}}}$

• Initial velocity of van (u) = 20m/s
• Final Velocity of van(v) = 60 m/s
• Time(t) = 5sec

$\: \huge{ \underline{ \large{ \rm{ \blue{Find:}}}}}$

• Displacement of the van?

$\: \huge{ \underline{ \large{ \rm{ \green{Solution:}}}}}$

$\: \: \: \: \: \: \: \: \large{ \boxed{ \rm{\orange{S = ut + ½ at²}}}}$

• This equation is known as the second equation of motion and it is used to calculate the distance travelled (s) by a body in time (t), the body having initial velocity (u) and acceleration (a).

So, let's find the acceleration:-

We know that ⤵

$\: \: \: \: \: \: \: { \boxed{\red{ \rm{v = u + at}}}}$

• where a is acceleration, v is the final velocity of the object, u is the initial velocity of the object and t is the time

⇒ v = u +at

⇒ 60 = 20 + a (5)

⇒ 60 = 20+5a

⇒ 5a = 60-20

⇒ 5a = 40

⇒ a = 8m/s²

⠀⠀⠀⠀⠀⠀⠀⠀

So, Acceleration of body is 8m/s²

⠀

☞Distance covered by van:-

⠀

S = ut + ½at²

S = 20 × 5 + ½×8 × 5²

S = 100 + ½ × 200

S = 100 + 100

S = 200m

⠀

$\boxed{ \therefore{ \sf{ \pink{Distance \: covered \: by \: van = 200m}}}}$