The velocity attained by the particle at the end of the time is calculated as,
v1=u+at1 =4t1 Therefore, the displacement in first
t1 seconds is given by,
s1=at12 =×4×t12 =2t12 The displacement in the next
t2 seconds is given by,
s2=v1t2 =4t1t2 For third part of journey,
Initial velocity is
v1=4t1, final velocity is 0 and time interval
t1 Thus, acceleration is given by,
a= =−4ms−2 Thus, displacement in the third part is,
s3= =2t12 Therefore,
v22−v12=2as3 Now, average velocity is calculated as,
vavg =| Total displacement |
| Total time |
5.1= 5.1= 51=4t12+4t1(10−2t1) Solve further.
4t12+40t1+51=0 (2t1−3)(2t1−17)=0 t1= or
t1= t1=1.5s As,
t1=8.5s is not possible. It exceeds total time.