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Question

A car starts from rest and moves along the x-axis with constant acceleration of 5m/s2 for 8seconds. If it then continues with constant velocity, what distance will the car cover in 12seconds since it started from the rest?


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Solution

Step 1: Given

Initial velocity, u=0

Acceleration, a=5m/s2

Time,

ttotal=12secondst1=8secondst2=4seconds

Step 2: Formula used

  1. The first equation of motion, v=u+at.
  2. The second equation of motion, s1=ut+12at2.
  3. Distance=Speed×time

Step 3: Determine the distance covered in the first 8seconds

Let be the distance travelled in the first 8seconds

Using the second equation of motion,

s1=ut+12at2=ut1+12at12=0(8)+12(5)(8)2=160ms1

Step 4: Determine the velocity at the end of 8 seconds when the car is accelerating

Using the first equation of motion,

v=u+at=u+at1=0+(5)(8)=40m/s

Step 5: Determine the remaining distance

The distance traveled in first 8seconds has been calculated.

The total time from the moment car starts from rest is 12seconds

Distance traveled in the last 4seconds will be s2

Distance=Speed×times2=v×t2=40×4=160m

Step 6: Determine the total distance

Totaldistance=s1+s2=160+160=320m

Therefore, the car covers 320m in total.


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