Starting from rest a scooter acquires …
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Starting from rest a scooter acquires a velocity of 36 km per hour in 10 seconds and then brakes are applied it takes 25 seconds to stop find out the acceleration and distance travelled during this motion
Posted by Mayank Som 1 year ago
- 2 answers
Kushagra Goyal 1 year ago
Case I
Initial velocity (u)=0m/s
Final velocity (v)= 36km/h=36×5/18= 10m/s [1km/h=5/18 m/s]
Time= 20 sec
Acceleration=?
By using first equation of motion
[V=u+at]
36 =0 + a×20
36 - 0 = 20a
36/20 =a
9/5 = a
1.8m/s²
Case II
Initial velocity (u)=36km/h= 36×5/18= 10m/s [1km/h=5/18 m/s]
Final velocity (v)= 0m/s
Time= 25 sec
Acceleration=?
By using first equation of motion
[V=u+at]
0= 36 + a×25
0-36=25a
-36/25=a
1.44m/s²
Case I
Initial velocity (u)=0m/s
Final velocity (v)= 36km/h=36×5/18= 10m/s [1km/h=5/18 m/s]
Time= 20 sec
Acceleration=1.8m/s²
Distance=?
By using second equation of motion
[s=ut+½at²]
0×20+½×1.8×20²
=360m
Case II
Initial velocity (u)=36km/h= 36×5/18= 10m/s [1km/h=5/18 m/s]
Final velocity (v)= 0m/s
Time= 25 sec
Acceleration=1.44m/s²
Distance=?
By using second equation of motion
[s=ut+½at²]
10×25+½1.44×25²
=700m
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Mohd Rihan 1 year ago
0Thank You