Question

A particle starts from rest moving along a straight line with a constant acceleration of 4 ms. the ratio of the distances travelled in the second and third second is
b) 100 m
b) 3:5
a) 3:7
d)9:5
c) 4:3​

Answer 1

Answer:

option (b) 3 : 5

Explanation:

Given :

A particle starts from rest moving along a straight line with a constant acceleration of 4 m/s².

To find :

the ratio of the distances travelled in the second and third second

Solution :

As the particle started from rest, initial velocity, u = 0 m/s

acceleration, a = 4 m/s²

The distance travelled in the nth second is given by,

 $\boxed{\bf S_n=u+a\bigg(n-\dfrac{1}{2} \bigg) }$

The distance travelled in the second second :

  $\sf S_2=0+4 \bigg (2-\dfrac{1}{2} \bigg) \\\\ \sf S_2=4\bigg( \dfrac{4-1}{2} \bigg) \\\\ \sf S_2=4 \times \dfrac{3}{2} \\\\ \sf S_2=6 \ m$

The distance travelled in the third second :

 $\sf S_3=0+4 \bigg (3-\dfrac{1}{2} \bigg) \\\\ \sf S_3=4\bigg( \dfrac{6-1}{2} \bigg) \\\\ \sf S_3=4 \times \dfrac{5}{2} \\\\ \sf S_3=10 \ m$

The ratio of the distances travelled in the second and third second is :

S₂ : S₃ = 6 : 10

S₂ : S₃ = 2(3) : 2(5)

S₂ : S₃ = 3 : 5

∴ The ratio of the distances travelled in the second and third second is 3 : 5

Answer 2

Answer:

3:5

Explanation:

A particle starts from rest moving along a straight line with a constant acceleration of 4 m/s².