a motorcycle accelerates uniformly from 54 km h to 72 km/h in 2s. calculate (i) the acceleration and (ii) the distance covered by the motorbike in that time
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1
Answer:
I) acceleration will be 2.5 m/s²
Step-by-step explanation:
2) distance covered will be 40 m as
speed of motor bike in metered is 20 m/s . It travels for 2 sec so distance covered in 2 sec is 20m + 20m = 40 m
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Initial velocity,
u = =54 km/h
= 54 (5/18) m/s
= 15 m/s
Final velocity,
v = 72 km/h
= 72 (5/18) m/s
= 20 m/s
Time in which velocity changes is, t = 2s
(a) Acceleration
Acceleration, a = (v - u)/t
a = (20-15)/2= 2.5 m/s^2
(b) Distance travelled by the bus
v² = u² + 2aS
20² = 15² + 2 × 2.5× S
400 = 225 + 5×S
400 - 225 = 5S
5S = 175 m
S = 35m
Distance = 35m
u = =54 km/h
= 54 (5/18) m/s
= 15 m/s
Final velocity,
v = 72 km/h
= 72 (5/18) m/s
= 20 m/s
Time in which velocity changes is, t = 2s
(a) Acceleration
Acceleration, a = (v - u)/t
a = (20-15)/2= 2.5 m/s^2
(b) Distance travelled by the bus
v² = u² + 2aS
20² = 15² + 2 × 2.5× S
400 = 225 + 5×S
400 - 225 = 5S
5S = 175 m
S = 35m
Distance = 35m
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