Question

An object moves with a uniform velocity of 18km/h ,for 5s. Then it starts accelerating at a uniform rate and attains velocity of 90km/h in next 20s. Find the

a) distance covered with uniform velocity.

b)acceleration of the body during the 20s

c) distance covered by body while accelerating

d) average speed during 25 s interval.
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❏ NOTE -:
NO SPAM
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QUESTION FOR MODERATORS​

Answer 1

Answer:

An object moves with a uniform velocity of 18km/h ,for 5s. Then it starts accelerating at a uniform rate and attains velocity of 90km/h in next 20s. Find the

a) distance covered with uniform velocity.

b)acceleration of the body during the 20s

c) distance covered by body while accelerating

d) average speed during 25 s interval.

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❏ NOTE -:

NO SPAM

QUALITY ANSWERS

QUESTION FOR MODERATORS

Answer 2

$\large\underline\mathtt \purple{ Question = > }$

An object moves with a uniform velocity of 18km/h ,for 5s. Then it starts accelerating at a uniform rate and attains of 90km/h in next 20s.

$\large\underline\mathtt \red{ Find \: the -: }$

→ a) distance covered with uniform velocity.

→ b)acceleration of the body during the 20s

→ c) distance covered by body while accelerating

→ d) average speed during 25 s interval.

$\large\underline\mathtt \green{ Answer = > }$

Given :—

→ uniform velocity = 18km/h

→ time = 5 seconds

→ uniform velocity in (seconds) (5) = 5m/s

so ,

→ uniform velocity = 5m/s

→ time will be = 5 seconds

After acceleration :→

→ uniform velocity = 90km/h

→ time taken = 20s

→ converting (km/h) into (m/s) =

then ,

→ uniform velocity = 25m/s

→ time taken = 20 seconds

Now,

$\bf \purple{ According \: To \: The \: Question }$

a) Distance covered with uniform velocity = ?

→ formula used = s = D/t

→ (speed) s = 5

→ (distance) d = ?

→ (time) t = 5 seconds

now putting the values in :—

→ 5 = D/5

→ 5×5 = D

→ 25m/s = D

∴ the distance covered by the object with uniform velocity is 25m/s.

b) acceleration of the body during 20s.

→ formula used = v=u+at

→ velocity after acceleration(v) = 25m/s

→ time = 20s

→ (u) = 0 (after acceleration)

now putting the values in :—

→ 25= 0+ (a× 20)

→ 25 = 0 + 20a

→ 25/20 = a

→ 1.25m/s² = a

∴ the acceleration of the body during 20s is 1.25m/s².

c) distance covered by the body while accelerating. = ?

→ formula used : s=ut+1/2at²

→ s (distance) = ?

→ u (Initial velocity) = 0

→ t (time) = 20s

→ a (Acceleration) = 1.25m/s². (done above)

now putting values in it :—

→ s = (0×20)+(1/2×1.25×20²)

→ s = 0+(1/2×1.25×20×20)

→ s = 1.25×10×20

→ s = 1.25×200

→ s = 250m/s

∴ distance covered while accelerating is 250m/s.

d) average speed during 25s interval. = ?

→ formula used : s = D/t

→ average speed (s) = ?

→ distance (d) = 250m/s

→ time (t) = 25 seconds

now putting values in it :—

→ s = 250/25

→ s = 10m/s²

∴ average speed during 25s interval is 10m/s².

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