Question

Starting from the rest a scooter acquines a velocity of 36km/h in 10sec and then breaks are applied it takes 20sec to stop calculate acceleration and distance travelled

Answer 1

initial velocity, (u) = 0m/s
time taken to acquire velocity(t1) = 10 s
v1 = 36km/hr = 10m/s
therefore, initial acceleration = v1/t1 = 1m/s^2

when the scooter stards retarding the motion
initial velocity (v1) = 10m/s
final velocity (v2) = 0m/s
time taken to stop = 20s
therefore,
retardation produced (a') = -(-10/20) = 0.5m/s^2
total distance travelled =
distance during 10s + diastance during next
20 s
$= \frac{1}{2}(1)( {10}^{2} ) + (10 \times 20) + \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{2} ( - 0.5) ({20}^{2} ) \\ = 50 + 200 - 100 = 150m$

Answer 2

Answer:

Initial velocity, (u) = 0m/s

time taken to acquire velocity(t1) = 10 s

v1 = 36km/hr = 10m/s

therefore, initial acceleration = v1/t1 = 1m/s^2

when the scooter stards retarding the motion

initial velocity (v1) = 10m/s

final velocity (v2) = 0m/s

time taken to stop = 20s

therefore,

retardation produced (a') = -(-10/20) = 0.5m/s^2

total distance travelled =

distance during 10s + diastance during next

20 s

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Explanation:

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