Question

7. An object accelerates from rest to a velocity 27.5 m/s in
10 sec. Then find distance covered by object in next 10
sec
(1) 550 m
(2) 137.5 m
(3) 412.5 m
(4) 275 m. plzzz somebody explain this​

Answer 1

Answer:

We know v=u+at

therefore a=(v-u)/t

here v=27.5m/s, u=0(since object starts from rest), t=10s

there for a=(27.5-0)/10=2.75m/s²

Distance covered in first 10 seconds:

s=ut+1/2at2

here u=0, a=2.75m/s2, t=10s

s=1/2at2 (because u=0)

s=1/2*2.75*(10)2

s=275/2=137.5m

Distance travelled in first 20s

s=ut+at2

here u=0, a=2.75m/s2, t=20s

s=1/2at2 (because u=0)

s=1/2*2.75*(20)2

s=550m

Distance required is the distance travelled from 11th to 20th seconds=Distance travelled in 20s - Distance travelled in 10s

i.e. 550-137.5=412.5m

Explanation:

please please mark as brainliest

Answer 2

Answer:

Hello❤

Initial velocity =0

Final velocity =27.5 m/s

Time=10 seconds

_______________________________

Using First law of Motion

v=u+at

(v-u)/t=a

(27.5-0)/10=27.5 n/s^2

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Using 2nd law of Motion

s=ut+0.5 at^2

(This time initial will be 27.5 m/s)

s=27.5*10+0.5*2.75*10^2

s=275+0.5*275

S=412.5m

__________________________

Thanks♡