Question

16. A particle starts rectilinear motion with initial velocity
5m/s under uniform acceleration 2 ms^2. How much
distance will be travelled by the particle is: 10s and
10th second​

Answer 1

u=5m/s

a=2m/s²

t=10s

By,s=ut+1/2at²,we get

s=5×10+1/2×2×(10)²

s=50+100=150m

Thus,distance travelled in 10s is 150m

Distance travelled in the nth second is

Sn=u+a/2(2n-1)

Sn=5+2/2(2×10-1)

Sn=5+19=24m

Thus,distance travelled in the 10th second is 24m

Answer 2

Answer:

10 seconds = 150 metres

10th second = 24 metres

Explanation:

Given :

• Initial velocity = u = 5 m/s

• Acceleration of the particle = a = 2 m/s²

• Time taken = t = 10 seconds

To find :

• Distance travelled by the particle in 10 seconds and distance travelled in 10th second

Using the second equation of motion :

S=ut+½at²

S=5×10+½×2×100

S=50+100

S=150 metres

Now distance travelled in 10th second:

Sn = u+a/2(2n-1)

Sn = 5+2/2(2×10-1)

Sn = 5+1(20-1)

Sn = 5+19

Sn = 24 metres

The distance travelled in 10 seconds is 150 metres and distance in 10th second is 24 metres