Question

Cars moving along a straight line at a speed of 54km/hr stop in 5s after the brakes are applied. (a) Find the acceleration, assuming it to be constant. (b) Plot the graph of speed versus time. (c) Using the graph. Find the distance covered by the car after the brakes are applied? 17 Acar moves100m due east and then 25m due west. (a) What is the distance covered by the car? (b) What is its displacement? 18 An athlete runs from one end to the other end of a semi circular track whose radius is 70m. What is the distance covered by the athlete and what is his displacement? [ Ans: Distance= 220m, Displacement= 140m] 19 A body starts from rest and moves with a uniform acceleration of 5m/s2 for 5s and then it moves with a constant velocity for 4s. Later it slows down and comes to rest in 5s. Draw the velocity graph for the motion of the body andanswerthe followingquestions: a. What is the maximum velocity attended by the body? b. What is the distance travelled during this period of acceleration? c. What is distance travelled when the body was moving with constant velocity? d. What is the retardation of the body while slowing down? e. What is the distance travelled by retarding? f. What is the total distance travelled? Ans: (a) 25m/s, (b) 6.25m, (c) 100m, (d) 5m/s2 , (e) 62.5m,(f) 225m 20 An athlete moves 150m in 2minutes and next 50m in 20s on the same straight path. What is his average speed and average velocity? 21 A swimmer swims 90m long pool. He covers the distance of 180m by swimming from one end to other end back along the same path. If he covers the first 90m at speed of 2m/s, then how fast he swim so that his average speed is 3m/s? 22 A vehicle moves at a speed of 40 km/h, It is stopped by applying brakes which produces a uniform acceleration of -0.6m/s2 . How much distance will the vehicle move before coming to stop? 23 Brakes applied to a car produces an acceleration of 5m/s2 in the opposite direction to the motion. If car takes 1.5s to stop after applying the brakes. Calculate the distance travelled by it? 24 A person walks along the sides of a square field each side is 100m long. What is the maximum magnitude of displacement of the person in any time interval? 25 Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of u12:u22 (Assume upward acceleration is –g and downward acceleration to be +g).

Answer 1

16.
u =54 kmph =54*5/18 =15m/s        t = 5s          v = 0
a = (v-u)/t = - 3 m/s
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v  =  u + a t = 15 - 3 t    m/s
table of points on graph.
t =     -1,      0,       1,     2,     3,     4,      5,     6    sec
v =    15,    15,    12,    9,     6,     3,       0,     0     m/s
scale on  v axis:    3m/s  per cm  or  1.5 m/s  per cm.    scale on  t axis :  1 sec per 1 cm.
distance = u t + a t^2 /2 = 15 * 5 - 3 * 5^2 / 2 = 37.5 m
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17.
Distance covered = 100 m + 25 m = 125 m
displacement = final position - initial positon = 75 m  = straight line distance of final position from the starting position.
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18. 
distance = semicircle perimeter = pi * 70 m = 22/7 * 70 = 220 m
Displacement = distance of final position from initial = diameter = 140 m
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19 
u = 0      a = 5 m/s^2       t = 5 s
v = u + a t = 25 m/s.         s = 1/2 a t^2 = 125/2 = 62.5 m
constant velocity = 25 m/s.    t = 4 sec.    s = v t = 100 m
now   u = 25 m/s       v =  0     t = 5 s.
a = (v-u)t = -5 m/s^2         s = u t + 1/2 a t^2  = 25 * 5 - 5 * 5^2 /2 = 62.5 m
Draw velocity vs time.
a. maximum velocity = 25 m/s
b. distance = 62.5 m
c. 100 m
d.   -5 m/s^2
e. 62.5 m
f. 62.6 + 100 + 62.5 = 225 m
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20. 
s1 = 150 m            t1 = 120 sec             v1 = s1/t1 = 1.25 m/s
s2 = 50m               t2 = 20 s                    v2 = 2.5 m/s
avg. speed = (s1+s2)/(t1+t2)  = 200/140 = 10/7   m/s
avg. velocity = avg. speed as direction of velocity remains same = 10 /7   m/s
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21.
s1=90m           s2 = 90 m       v1 = 2 m/s         t1 = 45s        t2 = ?           avg speed = 3 m/s
3 = (90+90) / (45 + t2)  
v2 = 15 m/s
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22.  u = 40 kmph = 100/9 m/s       a = - 0.6 m/s^2                v = 0
v^2 - u^2 = 2 a s 
s = 100*100/(9*9 * 2 * 0.6)  m  = 25,000/243   m
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23.
a = - 5 m/s^2            t = 1.5 s           v = 0
u = v - at = 7.5 m/s                  s = (v^2 - u^2)/2a  =  45/8  m
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24.       Let the person start from one vertex and traverse two sides to reach the other end of the diagonal.   At this point, the displacement is maximum = 100 * sqrt(2)  m
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25.    
u1,   u2.          v1 = 0     v2 = 0           a = - g
s1 = (v^2 - u^2)/2a = u1^2/2g
s2 = u2^2 /2g
Ratio =  u1^2 / u2^2