Question

A biker travelling on a straight road passes a traffic signal with a velocity of 4 m/s and accelerates uniformly for 5 seconds. Then he accelerates at double the rate for 3 more seconds and passes the second traffic signal with a velocity of 26 m/s. Find his speed 2 seconds after he passes the first traffic signal.

Answer 1

Answer:

Correct option is

A

0.712

Given, initial velocity, u=0m/s and final velocity v=7.10m/s

Distance traveled S=35.4m

If a be the acceleration.

Using formula v

2

−u

2

=2aS

(7.10)

2

−0

2

=2a(35.4)

⟹ a=0.712m/s

2

Answer 2

Answer:

Speed 2 seconds after he passes the first traffic signal = 30m/s.

Explanation:

Initially,

Initial velocity, $u_1=4m/s$

Let acceleration be denoted by $a$,

Time, $t=5s$

Let final velocity after 5 seconds be = $v_1$

Using newton's equation of motion, $v=u+at$

$v_1=u+at$

$\implies v_1 =4+5a$

Later,

Intial velocity = final velocity after 5 seconds = $v_1$

time, $t=3s$

Given: final velocity = $v_2 = 26m/s$

and acceleration = $2a$

Using newton's equation of motion, $v=u+at$

$v_2=v_1+(2a)t$

$\implies 26=(4+5a)+(2a)(3)$

$\implies 26=4+11a$

$\implies 22=11a$

$\implies a=2m/s$

For speed 2 seconds after he passes the first traffic signal:

initial velocity, $v_2=26m/s$

acceleration, $a=2m/s$

final velocity = $v_3$

time, $t=2s$

Using newton's equation of motion, $v=u+at$

$\implies v_3=26+(2)(2)$

$\implies v_3=30m/s$

Hence, speed 2 seconds after he passes the first traffic signal = 30m/s.

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