Q3) An object moves with a uniform velocity of 18km/h, for 5s. Then it starts accelerating at a uniform rate and attains velocity of 90km/h in next 20s. Find the a) distance covered with uniform velocity. b) acceleration of the body during the 20s c) distance covered by body while accelerating d) average speed during 25 s interval.
Answers
Answer:
Given :—
→ uniform velocity = 18km/h
→ time = 5 seconds
→ uniform velocity in (seconds) (5) = 5m/s
so ,
→ uniform velocity = 5m/s
→ time will be = 5 seconds
After acceleration :→
→ uniform velocity = 90km/h
→ time taken = 20s
→ converting (km/h) into (m/s) =
then ,
→ uniform velocity = 25m/s
→ time taken = 20 seconds
Now,
\bf \purple{ According \: To \: The \: Question }AccordingToTheQuestion
a) Distance covered with uniform velocity = ?
→ formula used = s = D/t
→ (speed) s = 5
→ (distance) d = ?
→ (time) t = 5 seconds
now putting the values in :—
→ 5 = D/5
→ 5×5 = D
→ 25m/s = D
∴ the distance covered by the object with uniform velocity is 25m/s.
b) acceleration of the body during 20s.
→ formula used = v=u+at
→ velocity after acceleration(v) = 25m/s
→ time = 20s
→ (u) = 0 (after acceleration)
now putting the values in :—
→ 25= 0+ (a× 20)
→ 25 = 0 + 20a
→ 25/20 = a
→ 1.25m/s² = a
∴ the acceleration of the body during 20s is 1.25m/s².
c) distance covered by the body while accelerating. = ?
→ formula used : s=ut+1/2at²
→ s (distance) = ?
→ u (Initial velocity) = 0
→ t (time) = 20s
→ a (Acceleration) = 1.25m/s². (done above)
now putting values in it :—
→ s = (0×20)+(1/2×1.25×20²)
→ s = 0+(1/2×1.25×20×20)
→ s = 1.25×10×20
→ s = 1.25×200
→ s = 250m/s
∴ distance covered while accelerating is 250m/s.
d) average speed during 25s interval. = ?
→ formula used : s = D/t
→ average speed (s) = ?
→ distance (d) = 250m/s
→ time (t) = 25 seconds
now putting values in it :—
→ s = 250/25
→ s = 10m/s²
∴ average speed during 25s interval is 10m/s².
Explanation:
I hope it will help you please make me as brainlist answer and please follow