The exercise will cover the topics acceleration uniform and non uniform it will help you to get the concept of the acceleration and when it is uniform and non uniform
Intext Exercise 1.5
Que1: A bus starting from rest moves with uniform acceleration of 0.1m/s² in 2minutes. Find out(i) speed acquired
(ii) the distance travelled
Ans1: (i) Given
Time interval, t = 2* 60= 120 sec
Initial velocity u=0
Acceleration, a= 0.1 = v-u/t
0.1 = v – 0 /120
120*0.1 = v
12 m/s = v
(ii) Given
Final Velocity, v = 12m/s
Initial velocity, u= 0
Acceleration, a= 0.1m/s²
Time interval, t = 120 sec
Distance travelled, s =?
According to the third equation of motion v²- u²= 2as
Putting the values
(12²) – (0²) = 2*0.1*s
144/0.2 = s
720 m = s
The speed acquired is 12m/s and distance is 720m
Que2:A train is travelling at uniform speed of 90km/h. Breakes are applied to get uniform acceleration of – 0.5 m/s²
Find how far train will go before it is brought to rest.
Ans2: Given
Speed, u = 90km/h = 90*5/18 = 25m/s
Acceleration, a = -0.5m/s²
Final velocity, v = 0
Distance covered before it is brought to rest, s=?
According to the third equation of motion v² – u² = 2* a* s
Putting all the values
0 – ( 25²) = 2 *(-0.5)* s
– 625/-1 = s
S = 625 m
Que3: A trolley while going from inclined plane has acceleration of 2cm/s² what will be the velocity 3s after it starts?
Ans3 : initial velocity, u= 0
Given acceleration, a= 2 cm/s² = 0.02 m/s²
Time interval = 3sec
According to v= u + at
v = 0+0.02*3
v = 0.06 m/s
Que4:A car racing at uniform acceleration of 4m/s². What will be the distance covered in 10s after start.
Ans4: Given
Acceleration, a= 4m/s²
Time interval, t = 10s
Initial velocity, u= 0
According to equation 2 of motion
S = ut + 1/2at²
S = 0*10 + 1/2 *4*(10²)
S = 200m
Que5: A stone is thrown vertically upward direction with velocity of 5m/s
If the acceleration of stone during its motion is 10m/s² in downward direction what will be the height attained by stone and how much time it will take to reach there.
Ans5:
Given initial velocity, u= 5 m/s
Final velocity, v= 0
Acceleration,a= – 10m/s² ( the direction of acceleration is downward that is opposite to velocity that’s why it is in negative)
According to v= u + at
0 = 5 +(-10)t
5/10 => 0.5sec = t
According to the third equation of motion v² – u² = 2* a* s
Putte values
(0²)- (5²) = 2*(-10)* s
( – 25) / (- 20) = s
Height = 1.25m
