A motercar with initial velocity 22 km/h attains velocity 94 km/h in 1/12 hour. If acceleration is uniform then find (i) Value of acceleration of car and (ii) Total distance travelled by car for attaining this velocity.
Answers
Answer:
Given u=36 km/h=1836×5 m/s
u=10 m/s
v=?
t=1 sec
a=5m/s2
From 1st equation of motion
v=u+at
v=10+5(1)
v=15
Final velocity, v=15 m/s .
Given :
Initial velocity = 22 kmph
Final velocity = 94 kmph
Time interval = 1/12 hr
To Find :
Acceleration of car and distance travelled by car in the given interval of time.
Solution :
❖ Acceleration is defined as the rate of change of velocity.
- It is a vector quantity having both magnitude as well as direction.
- SI unit : m/s²
First of all we have to convert unit of velocity from kmph to mps
We know that, 1 kmph = 5/18 mps
• u = 22 kmph = 22 × 5/18 = 6.11 mps
• v = 94 kmph = 94 × 5/18 = 26.11 mps
Time = 1/12 hr = 3600/12 = 300 s
➙ v = u + at
- v denotes final velocity
- u denotes initial velocity
- a denotes acceleration
- t denotes time
➙ 26.11 = 6.11 + 300a
➙ 20 = 300a
➙ a = 20/300
➙ a = 0.067 m/s²
♦ Distance covered by car :
Applying 3rd equation of kinematics;
➠ v² - u² = 2as
➠ (26.11)² - (6.11)² = 2(0.067) × s
➠ 681.73 - 37.33 = 0.134 s
➠ s = 644.4 / 0.134
➠ s = 4809 m