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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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
The final velocity = v= 65 ms-¹
Distance covered = S = 200 m
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