a motorbike initially at rest, picks up a velocity of 72km/hr over a distance of 40m. Calculate (i) acceleration (ii) time in which it picks up above velocity
Answers
Given that, a motorbike initially at rest (initial velocity i.e. u is 0 m/s), picks up a velocity of 72km/hr (final velocity i.e. v is 72 km/hr) over a distance of 40m (s is 40 m).
We have to find the acceleration (a) and time (t).
a) For acceleration:
→ v² - u² = 2as
The final velocity is in km/hr. So, convert it in m/s.
= 72 × 5/18 = 20 m/s
Substitute the known values
→ (20)² - (0)² = 2 × a × 40
→ 400 - 0 = 80a
→ 400 = 80a
→ 400/80 = a
→ 5 = a
Therefore, acceleration of the motorbike is 5 m/s².
b) For time:
→ v = u + at
Substitute the known values,
→ 20 = 0 + 5(t)
→ 20 = 5t
→ 20/5 = t
→ 4 = t
Therefore, the time taken by motorbike is 4 second.
Answer:-
Given :
Initial Velocity (u) = 0 m/s.
Final Velocity (v) = 72 * 5/18 m/s = 20 m/s.
Distance travelled (S) = 40 m.
We know that,
v² - u² = 2*a*S ( 2nd equation of motion)
→ (20)² - (0)² = 2*a*(40)
→ 400 = a*80
→ a = 5 m/s² (1)
Now,
Using (1)st equation of motion,
→ v = u + at
→ 20 = 0 + (5)t
→ 20 = 5*t
→ 20/5 = t.
→ t = 4 s (2).
(or)
Using 3rd equation of motion,
→ S = ut + 1/2*at²
→ 40 = (0)(t) + 1/2 * 5 * t²
→ 40 * 2/5 = t²
→ 16 = t²
→ t = ± 4
As time cannot be negative , positive value is taken. i.e., 4 s.(2)