Question

A bus is travelling at speed of 15m/s. Then it accelerates for 5s at a rate of 10m/s2. Find the final velocity and distance covered by the bus during this process.



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Answer 1

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Given,

Initial velocity = u = 15 ms-¹

Acceleration = a = 10 ms-²

Time taken = t = 5s

Now,

Using first equation of motion we can find the final velocity,

v = u + at

v = 15 + 10 × 5

v = 65 ms-¹

Again,

we can solve the distance covered by using

we can solve the distance covered by usingboth second and third equation of motion.

S = ut + 1/2 × a × t²

S = 15 x 5 + 1/2 × 10 × 5 × 5

S = 75 + 125

S = 200m

Using third equation,

v ^ 2 - u ^ 2 = 2as

(v - u)(v + u) = 2(10)s

[v² - u² = (v-u)(v+u)

Hence

(65 - 15)(65 + 15) = 205

20S = 400C

S = 200m

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The final velocity = v= 65 ms-¹

Distance covered = S = 200 m