Question

A particle starts moving from rest state along a straight line under the action of a constant force and travel distance x in first 5 sec. The distance travelled by it in next five seconds will be

Answer 1

Distance travelled in first 5 seconds = x
S = ut + 0.5at²
x = 0 + 0.5a(5²)
x = 25a/2

Distance travelled in first 10 seconds = y
S = ut + 0.5at²
y = 0 + 0.5a(10)²
y = 100a/2

Distance travelled in next 5 seconds = y - x = 100a/2 - 25a/2 = 75a/2
75a/2 = 3 ×(25a/2) = 3x

Distance travelled in next 5 seconds is 3x

Answer 2

Answer:

Explanation:

Solution,

Here, we know that

According to the 2nd equation of motion,

s = ut + 1/2 at²

Distance travelled in first 5 seconds = x,

So, putting all the values, we get

s = ut + 0.5at²

x = 0 + 0.5a(5²)

x = 25a/2

And for the other,

Distance travelled in first 10 seconds = y

s = ut + 0.5at²

y = 0 + 0.5a(10)²

y = 100a/2

Distance travelled in next 5 seconds = y - x

= 100a/2 - 25a/2

= 75a/275a/2

= 3 ×(25a/2)

= 3x

Hence, the distance travelled in next 5 seconds is 3x.