question_answer

A car accelerates from rest at a constant rate for some time and attains a velocity of 8 m/s. Afterwards, it decelerates at a constant rate and comes to rest. If the total time taken is 4 s, find the total distance travelled by the car.

Answer:

Let\[{{t}_{1}}\]and\[{{t}_{2}}\]be the times of acceleration and retardation respectively. Acceleration part: \[u=0,\,v=8m/s,\,\,a=+a,\,\,t={{t}_{1}}\] We know,\[v=u+at\] \[\Rightarrow 8=a{{t}_{1}}\,\,or\,\,{{t}_{1}}=\frac{8}{a}\]                                             ……(1) Retardation part: \[u=8m/s,\,\,v=0,\,\,a=-a,\,\,t={{t}_{2}}=4-{{t}_{1}}\] We know,\[v=u+at\] (or) \[0=8-a(4-{{t}_{1}})\] \[\therefore 8=a\left( 4-\frac{8}{a} \right)\]                                        [from (1)] \[8=4a-8\,\,or\,\,a=4\,\,and\,\,{{t}_{1}}=8/4=2s\] Now, \[{{s}_{1}}=0\times 2+\frac{1}{2}\times 4{{(2)}^{2}}\,\,or\,\,{{s}_{1}}=8m\] \[{{s}_{2}}=8\times 2+\frac{1}{2}\times 4{{(2)}^{2}}\,\,or\,\,{{s}_{2}}=8m\] \[\therefore {{s}_{1}}+{{s}_{2}}=16m.\]