A car starts from rest and moves along the x axis with constant acceleration 5 M per second square for 8 seconds if then continues with constant velocity what distance will the car cover in 12 seconds since it started from rest
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Here,
u=0 m/s
a=5m/s^2
t=8sec
Applying s=ut+1/2at^2,we get
s=1/2*5*8*8
=160m
Now, final velocity after 8 secs of start
Applying v=u+at
v=0+5*8
v=40m/s
The car travels in constant velocity for (12-8)sec=4sec
Distance covered in the remaining 4 sec, applying s=ut+1/2at^2
Here, u=40m/s
a=0
t=4sec
s=40*4
=160m
Final distance covered by the car=160m + 160m
=320m
Thus, the answer is 320m
u=0 m/s
a=5m/s^2
t=8sec
Applying s=ut+1/2at^2,we get
s=1/2*5*8*8
=160m
Now, final velocity after 8 secs of start
Applying v=u+at
v=0+5*8
v=40m/s
The car travels in constant velocity for (12-8)sec=4sec
Distance covered in the remaining 4 sec, applying s=ut+1/2at^2
Here, u=40m/s
a=0
t=4sec
s=40*4
=160m
Final distance covered by the car=160m + 160m
=320m
Thus, the answer is 320m
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0
Answer:
320 m is the answer
Step-by-step explanation:
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