Question

A motor car takes 10 seconds to cover 30 meters and 12 seconds to cover 42 meters. Find theuniform acceleration ​

Answer 1

As given the , the motor car takes 10 seconds to cover 30m and 12 second to cover 42m, then we have to find uniform acceleration.

Let u is initial velocity and a is acceleration of motor car.

using formula, S= ut + 1/2at^{2}

1)t =10s, s=30m

so, 30m= u(10)+1/2a(10)^{2}

30 = 10u +50a

5= u+5a...........1

velocity after 10s v=u +a(10)=I+10

2) t=12s, s=42s

so, 42=v(12)+1/2a(12)^{2}

42=12(u+10a) +1/2a(144)

7=2u +32a.............2

solving equations 1 and 2

eq2 -eq2×1

7 - 2×3=2u +32a=2(u+50)

1=2u+32a-2u-10a

1=22a

a=1/22=0.04545m/s^{2}

hence the acceleration is 0.04545m/s^{2}

Answer 2

Answer:

Case I

Distance covered by car =30 metres

Time taken = 10 second

So as velocity =distance /time

velocity =30/10 = 3m/sec

Case II

Distance travelled =42 metres

Time taken =12 seconds

Then, velocity =distance /time

velocity =42 /12 =7/2=3.5 m/sec

Also, change in time =(12-10)seconds =2 seconds

Now as, Acceleration =[change in velocity /change in time]

Acceleration =(3.5-3)/2 ms¯²

Acceleration =0.5/2 ms¯²

Acceleration =0.25 ms¯².

Now, by using the first equation of motion,

v=u+at

Note :Velocity after 15th second =velocity before 16 seconds.

Then,

v = 0 +( 0.25)×(16)

v = 4 m/s.

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