Question

A motorcycle moving with a speed of 5m/s is subjected to an acceleration of 0.2m/s. Calculate the speed of the motor cycle after 10 sec,and the distance travelled in this time..
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Answer 1

initial speed of car=5 m/s

 

accl=0.2m/s2

 

time=10s

 

distance=?

 

 

 

solution: s= ut+1/2at2

 

              s=(5)(10)+1/2 (0.2)(10)(10)

 

                  (50)+1/2(20)

 

                  =(50)+(10)

 

                  =60m

 

 

 

There’s an alternate method too:

 

u = 5 m/s

 

a = 0.2 m/s²

 

t = 10 seconds

 

v = u + at

 

= 5 + 0.2×10

 

= 5+2

 

= 7 m/s

 

s = ut + 1/2at²

 

= (5×10) + (1/2)×(2/10)×(10×10)

 

= 50 + 10

 

= 60 m

 

Final speed = 7 m/s

 

Distance travelled in 10 seconds = 60 m

 

Even this ones nice method:

 

 

u = 5 m/sec

 

a = 0.2 m/sec ^ 2

 

t = 10 sec

 

USING 1ST EQ OF MOTION,

 

v = u + at

 

v = 5 + 0.2×10

 

v = 5 + 2

 

v = 7 m/sec

 

 

 

FINAL VELOCITY = 7 m/sec

 

 

 

2nd EQ OF MOTION,

 

s = ut + 1/2 at^2

 

s = 5 × 10 + 1/2 × 2/10×10×10

 

s = 50 + 10

 

s = 60m

 

 

 

DISTANCE = 60 m

 

 

 

I have provided uh with 3 methods, plz follow up with ur convinient one ;)

 

 

 

Hope This Helps :)

Answer 2

Answer:

Explanation:

Initial velocity, u=5m/s Final velocity, v=? Acceleration, a=0.2m/s2 Time, t=10 sec Using , v=u + at v=5 + 0.2 x 10 v=5 + 2 =7 m/s Now distance travelled in time is calculated-acceleration-of-0-2-m-s-2