A car starting from rest accelerates uniformly with 5m/s^2 for sometime and then decelerates to come to rest with 3m/s^2. Find the maximum velocity attained during motion and the distance covered in a total time of 6 seconds.

Dear Student,

Please find below the solution to the asked query:

Let the distance travelled is S, first the car is travelling with acceleration a equals 5 space bevelled m over s squared. Let this acceleration is for time t1 and the car decelerates for time t2. Then,

V equals a t subscript 1 rightwards double arrow t subscript 1 equals V over 5 V equals 3 t subscript 2 rightwards double arrow t subscript 2 equals V over 3 T h e r e f o r e comma t subscript 1 plus t subscript 2 equals 6 space s rightwards double arrow V over 5 plus V over 3 equals 6 rightwards double arrow fraction numerator 3 V plus 5 V over denominator 15 end fraction equals 6 rightwards double arrow 8 V equals 90 rightwards double arrow V equals 90 over 8 rightwards double arrow V subscript m a x end subscript equals 11.25 space bevelled m over s

So, the total distance travelled is,

S subscript 1 equals fraction numerator V squared over denominator 2 a end fraction equals fraction numerator 126.6 over denominator 2 cross times 5 end fraction rightwards double arrow S subscript 1 equals 12.7 space m S subscript 2 equals fraction numerator V squared over denominator 2 a end fraction equals fraction numerator 126.6 over denominator 2 cross times 3 end fraction rightwards double arrow S subscript 1 equals 21.1 space m S o comma space t o t a l space d i s tan c e space i s comma S equals S subscript 1 plus S subscript 2 equals 12.7 plus 21.1 rightwards double arrow S equals 33.8 space m

 

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