Question

A scooter is travelling at a speed of 36 km/h. After one second, its velocity becomes 45 km/h. Calculate the acceleration experienced by the scooter and the distance covered by it in one second.

Answer 1

Answer:

s=distance; t=time; a = acceleration

a = (v2-v1)/ (t2-t1)

v1= 36 km/h

v2=45 m/s *3600/1000= 162 km/h

a = (162 km/h - 36 km/h)/1s * 1000/3600 = 35 m/s²

s = a/2 * t² + v(0) * t + s(0) ;

s = (35m/s² * 1²s²)/2 = 17,5 M

ADD IT AS A BRAINLIST ANSWER

Answer 2

Initial Velocity (u) = 36 km/h = 10 m/s

Final Velocity (v) 45 km/h = 12.5 m/s

Time (t) = 1 s

Acceleration (a) = ?

Distance (s) = ?

a) Using v = u + at

12.5 = 10 + a × 1

Acceleration (a) = 2.5 m/s²

b) Using s = ut + 1/2 at²

10 × 1 + 1/2 × 2.5 × (1)²

10 + 1.25

Therefore, the distance covered is = 11.25 m