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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Answer 1

Explanation:

Given parameters

Time taken (t) = 5 sec

Initial velocity (u) =18 km/hour

\(u = \frac{18\times 1000}{60\times 60s}\)

u = 5 m/s

Final velocity (v) =36km/hour

\(v = \frac{36\times 1000}{60\times 60s}\)

v =10 m/s

We need to calculate acceleration and distance traveled

We know that

acceleration a = (v – u)/t

Substituting the given values in the above equation we get,

a = (10 – 5)/5

a =5/5

a = 1 m/s2

We know that distance travelled is calculated by the formula

Distance travelled S = u t + (1/2) a × t2

Substituting the given values in the above equation we get,

S = 5 × 5 + (1/2) × 1 × 52

S = 25 + (1/2) × 25

S = 25 + 12.5

S = 37.5 m

Hence

acceleration a =1 m/s2

Distance travelled S =37.5 m

=10ms

−1

hope it will be help you

Answer 2

average speed during 25