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​

Answer 1

Answer:

412.5 m

Explanation:

As it was at rest, where velocity was 0,

Initial velocity = u = 0 m/s

Final velocity = v = 27.5 m/s

Time taken = t = 10 s

Using equations of motion:

= > v = u + at

= > 27.5 = 0 + a(10)

= > 27.5/10 = a

= > 2.75 m/s² = a

After 10 second:

= > v = u + at

= > v(after 10 sec) = present velocity + at

= > velocity after 10 sec = 27.5 + (2.75)(10) = 55 m/s

= > v² = u² + 2aS

= > (55)² = (27.5)² + 2(2.75)S

= > 3025 - 756.25 = 5.5S

= > 2268.5 = 5.5S

= > 412.5 m = S

Option (3)

Answer 2

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.

. ☞ options (d) ☜

.

.

.

Initial velocity =0

Final velocity =27.5 m/s

Time=10 seconds

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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

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