Question

A car enters a tunnel with velocity 4 metre per second moves with constant acceleration and comes out of the tunnel with velocity 8 metre per second velocity of the car at the middle of the tunnel is

Answer 1

Answer:

u =4m/s , a= 1 m/s² （given）

S = ut+0.5at² …..

displacement in 10 sec

S1 = 4 * 10 +0.5 *1 *100 = 90 meter

displacement in 9 sec

S2 = 4 * 9 + 0.5 * 1 *81 = 76.5 meter

displacement in 10th sec （last sec）

S = S1 - S2

S = 90 - 76.5 = 13.5 meter

Answer 2

In the case of uniform acceleration, there are three equations of motion which are also known as the laws of constant acceleration. Hence, these equations are used to derive the components like displacement(s), velocity (initial and final), time(t) and acceleration(a). Therefore they can only be applied when acceleration is constant and motion is a straight line. The three equations are,

v = u + at
v² = u² + 2as
s = ut + ½at²
where, s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time of motion. These equations are referred as SUVAT equations where SUVAT stands for displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time ().

Velocity of car at starting of tunnel Va = 4 m/s
Velocity of car after coming out of tunnel Vb= 8 m/s

We have to find velocity of car at ths the middle of the tunnel. Vc = ? m/s


Given velocity of the car at point A v=4 m/s and velocity of the car at point b u distance betweenA and B = s
From equation of motion we get,
v² = u² + 2as
(8)² = (4)² + 2as
64 - 16 = 2as
48 = 2as -------------(1)
Now, at the midpoint of the tunnel x= s/2
v² = u² + 2as/2
v² = (4)² + 24
v²= 40
v = √40
v = 6.32 m/s
6.32 /
# 3