Question

A truck is moving at a speed of 50 ms-1
Find stopping distance if it retards at a rate of 20 m/min.

Answer 1

Explanation:

ANSWER

Avg speed × total times = total distance

20×25=500

Since acceleration in the 1st interval is same as deceleration in the 3rd interval and deceleration is from the same velocity achieved after acceleration, t

1

=t

3

s

1

=

2

1

×5t

1

2

s

2

=(5t

1

)t

2

s

3

=(5t

1

)×t

3

−

2

1

×5×t

1

2

=(5t

1

)×t

1

−

2

1

×5×t

1

2

s

1

+s

2

+s

3

=5t

1

t

2

+5t

1

2

⇒5t

1

2

+5t

1

t

2

=500−−−−1

2t

1

+t

2

=25−−−−2

Solving (1 ) and (2) ⇒t

1

=5,t

2

=15sec