Question

plz answer my question.

Answer 1

I hope this help you

Answer 2

Let's assume that the maximum speed it attains is '"s"

Now distanve travelled during accelration+distance travelled durin decellaration=1500m
$\frac{ v^{2} - u^{2} }{2a} + \frac{ v^{2} - u^{2} }{2[-a]} =1500m$
$\frac{ s^{2} - 0^{2} }{2*5} + \frac{ 0^{2} - s^{2} }{2[-10]} =1500m$
$\frac{ s^{2} }{10} + \frac{ s^{2} }{20} =1500m$
$\frac{ 3s^{2} }{20} =1500m$
$S^{2}=10000m/ s^{2}$
$S=100m/ s^{2}$

time taken to attain this speed by accelration
$t= \frac{v-u}{a} = \frac{100-0}{5}=20s$

time taken to attain this speed by deaccelration
$t= \frac{v-u}{2a}= \frac{0-100}{2[-10]}=5s$

total time
=20+5=25s