UPSEE 2017 Solved Paper

We have,
x(t)=5t2−7t+3 ........(i)
By differentiating both side w.r.t. ' t ', we get
Velocity of the particle, v=

dx

dt

=10t−7 .......(ii)
We have to find acceleration at the moment, when velocity becomes 5m∕sec
∴ From Eq. (ii), we get
10t−7=5
∴t=

12

10

Now, on differentiation both side of Eq. (ii) w.r.t. ' t ', we get acceleration of the particle.
∴ a=

dv

dt

=10m∕sec2
Here, acceleration is constant all the time.
∴a=

dx

dt

|t=

12

10

=10m∕sec2