7.A train travels with a speed of 60km/h from station A to station B and then comes back with a speed 80km/h from station B to station A. Find a) the average speed
b)the average velocity of train
Ans.Speed from A to B = 60km/h
Speed from B to A = 80 km/h
Take the LCM of 60 & 80 = 240
Take it as distance between station A & B.
s₁=60km/h
d₁=240 km
t₁=?
S=d/t
⇒S×t=d
⇒t=d/s
t₁=d₁/s₁
⇒t₁=240km /60kmh-¹
⇒t₁= 4h
s₂=80kmh-¹
d₂=240km
t₂=?
S=d/t
t₂=d₂/t₂
⇒t₂=240km /80kmh-¹
⇒t₂=3h
Average speed=total distance / total time taken
⇒Average speed=( d₁ + d₂)/(t₁+t₂)
⇒average speed=(240+240)km / (4+3)h
⇒average speed=480km / 7
⇒average speed=68.571 kmh-¹
Average velocity=total displacement / total time taken
Total displacement is zero because body the body returns back to its initial position.
Average velocity= 0 / 7h
=0 km / 7h
= 0 kmh-¹
8. A train is moving with a velocity of 90 km per hour. It is brought to stop by applying the brakes which produce a retardation of 0.5 metre per second square. Find: a) the velocity after 10 second b) the time taken by the train to come to rest.
Ans.u=90km/h
u=90×1 kmh-¹
u= 90×5/18 ms-¹
u= 5×5 ms-¹
u= 25 ms-¹
a= - 0.5ms-²
v=?
t= 10s
v= u+ at
v= 25ms-¹ + (-0.5ms-²)(10s)
v=25ms-¹ - 5ms-¹
v= 20 ms-¹
For the second part,
Body comes to rest, it mean's final velocity v is zero.
u=25 ms-¹
v=0 ms-¹
t=?
a= 0.5 ms-²
v² - u² = 2as
(v² - u²)/2a = s
s= (v² - u²) / 2a
s= {( 0)² - (25 ms-¹)²} / 2×-0.5ms-²
s= - 25× 25 m²s-² / - 1.0 ms-²
s= 625 m.
v= u + at
v- u = at
(v - u )/ a = t
t = (v- u) / a
t= (0 - 25 ms-¹)/ - 0.5 ms-²
t= - 25 ms-¹ / - 0.5 ms-²
t= 50 s.
9.a car travels a distance hundred metre with a constant acceleration and average velocity of 20 metre per second. The final velocity acquired by the car is 25 metre per second. Find the initial velocity and acceleration of car?
Ans. s= 100 m
Average velocity=20 ms-¹
Final velocity (v)=25 ms-¹
u=?
a=?
Average velocity=( initial velocity +final velocity)/ 2
20ms-¹ = (u + v)/2
20 ms-¹ × 2 = ( u + 25 ms-¹ )
40ms-¹ - 25 ms-¹ = u
15 ms-¹ = u
u = 15 ms-¹
But acceleration is equal to change in velocity divided by time but that formula cannot be used because time has not been given to us.so,
v² - u² = 2as
a= (v² - u²)/ 2s
a= {(25ms-¹)² - (15ms-¹)²}/ 2×100m
a= {625 m²s-² - 225m²s-²}/200m
a= 400m²s-² / 200 m
a= 2 ms-²
10. When brakes are applied to a bus,the reservation produced is 25 cm per second square and the bus takes 20 seconds to stop. Calculate: the initial velocity of bus, and the distance travelled by bus during this time?
Ans. a= - 25 cm s-²
a= - 0.25 ms-². (1cm = 1/100 m)
t= 20s
v= 0 ms-¹. (bus comes to rest)
u=?
s=?
v= u + at
v- u =at
v - at = u
u= v - at
u= 0 - ( -0.25 ms-²)× 20s
u= + 5.0 ms-¹
S= ut + 1/2 at²
s= 5ms-¹ × 20s + 1/2 × (-0.25 ms-²) × 20s×20s
s= 100m - 0.25 ms-²× 200s²
s= 100m - 50m
s= 50m.
11. A body moves from rest with uniform acceleration and travels 270 metre in 3 seconds. Find the velocity of the body at 10 seconds after the start?
Ans. u=0 ms-¹
s= 270 m
t= 3s
v=?
s= ut + 1/2 at²
S- ut = 1/2 at²
2(s - ut) = at²
2(s - ut)/ t² = a
a= 2(s - ut)/ t²
a= 2( 270 m - 0× 3s)/ (3s)²
a= 2× 270m / 9s²
a= 2× 30ms-²
a= 60 ms-²
v= u+ at
v= 0 + 60ms-² × 10s
v= 600ms-¹.
12.A body moving with a constant acceleration travels the distance 3 and 8 respectively in one second and two seconds. Calculate the initial velocity, and acceleration of body.
Ans. S₁= 3m
t₁=1s
s₂=8m
t₂=2s
u=?
a=?
s=ut + 1/2 at²
s₁=ut₁ + 1/2 a(t₁)²
3 = u×1 + 1/2 a× 1×1
3= u + 1/2 a .........Ⅰ
s=ut + 1/2 at²
s₂= ut₂ + 1/2 a(t₂)²
8= u×2 + 1/2 a × 2×2
8= 2u + 2 a .............Ⅱ
Solving equation Ⅰ and equation Ⅱ we get,
Solving by the elimination method of linear equation in two variables.
Multiplying equation one by 2,
2(3=u + 1/2 a)
6 = 2 u + a. .......Ⅲ
Subtracting equation Ⅲ and Ⅱ , we get
8 - 6 =( 2u + 2a) - (2u + a)
2 = a
a= 2.
Unit of acceleration is metre per second square.so,
a= 2 ms-²
Not putting the value of a =2 in equation 3 we get,
6= 2u + a
6 = 2u + 2
6- 2 = 2u
4 = 2u
4/2 = u
2 = u
u = 2
Unit of velocity is metre per second.
so,
u = 2 ms-¹
13. A car travels with a uniform velocity of 25 metre per second for 5 seconds. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 seconds. Find:the distance with the car travels before the brakes are applied and the distance travelled by the car after applying the brakes and the retardation?
Ans. u=25ms-¹
t=5s
S₁=?
Distance travelled=velocity ×time
s₁=u × t
s₁=25ms-¹ × 5s
=125m
Retardation in 10 seconds:
u=25ms-¹
t=10s
v=0ms-¹
a=?
v= u + at
v- u =at
(v- u)/t= a
a= (v- u)/t
a=(0 - 25ms-¹)/10s
a= - 25ms-¹ / 10 s
a= - 2.5 ms-²
Reatardation= 2.5 ms-²
The distance travelled by the car after applying the break:
u= 25ms-¹
t=10s
a=-2.5ms-²
s₂= ?
S=ut +1/2 at²
S₂= 25ms-¹ × 10 s + 1/2 ×(-2.5ms-²) × 10s×10s
=250m - 2.5×5×10m
= 250m - 125m
= 125m
14.a spacecraft flying in a straight course with a velocity of 75 kilometre per second fires its rocket motors for 6.0 seconds. At the end of this time, its speed is 120 kilometre per second in the same direction. Find:
a) the spacecrafts average acceleration while the motors were firing,
b)the distance travelled by the spacecraft in the first ten seconds after the rocket motors was started, the motors having been in action for only 6.0 seconds.
Ans.u=75km/s
v=120km/s
t=6.0s
a=?
a=(v- u)/t
a=(120km/s - 75km/s)/ 6s
=45 kms-¹ / 6s
= 7.5kms-²
Distance travelled by spacecraft in in first 6 seconds and then in 4 seconds:
u=75kms-¹
t=6s
a=7.5kms-²
s₁=?
s= ut + 1/2 at²
s₁=75kms-¹ × 6s + 1/2 × 7.5kms-² × 6s × 6s
= 450km + 7.5×3×6km
= 450km + 135km
= 585 km
S= v×t
v=120kms-¹
t=4s
s₂=v× t
= 120 kms-¹ × 4s
= 480km
Total distance travelled= s₁ + s₂
=585km + 480 km
= 1065km
15.a train starts from rest and accelerates uniformly at a rate of 2 metre per second square for 10 seconds. It then maintains a constant speed for 200 seconds. The brakes are then applied and the train is uniformly retarded and comes to rest in 50 seconds. Find:
a) the maximum velocity reached
b) the retardation in the last 50 seconds.
c) the total distance travelled and
d) the average velocity of the train.
Ans.u=0ms-¹
a=2ms-¹
t=10s
v=?
v=u + at
v= 0ms-¹ + 2ms-² × 10s
= 0 + 20ms-¹
= 20ms-¹
Retardation in the last 50 seconds:
u=20ms-¹
v=0 ms-¹ (car comes to rest after 15 seconds)
t=50s
a=?
a= (v- u )/ t
a= (0 - 20ms-¹) / 50s
a= - 20ms-¹ / 50s
a= - 0.4ms-²
Total distance travelled:
Distance travelled in first 10 seconds
s₁=ut + 1/2 at²
= 0×10s + 1/2 × 2ms-² ×10s × 10s
= 0 + 100m
=100m
Distance travelled with a constant speed During 200 seconds,
s₂=v×t
= 20ms-¹ × 200s
= 4000m
Distance travelled in 50 seconds just before coming to rest,
s₃=ut + 1/2 at²
=20ms-¹ × 50s + 1/2 × (-0.4ms-²)×50s ×50s
= 1000m - 0.2ms-² ×2500s²
= 1000m - 500m
=500m
S=s₁+ s₂ +s₃
= 100m + 4000m + 500m
= 4600m
Average velocity of the train:
Average velocity=total displacement/total time taken
= 4600m / 260s. (Total time=10s+200s+50s)
= 17.69 ms-¹
No comments:
Post a Comment
Thanks