Question

1 point) A truck starts from rest and accelerates at
1?/?^{2}. 6 s later, a car accelerates from rest at the
same starting point with an acceleration of
2.7?/?^{2}.

a) Where and when does the car catch the truck?

1. ?m from the starting point

2. at ? seconds from the moment the truck started to
accelerate.

b) What are their velocities when they meet?

The truck : m/s

The car : m/s

Answer #1

Initial velocity of Truck,V_{iT}=0

Initial velocity of car,V_{ic}=0

Acceleration of truck,a_{T}=1 m/s^{2}

Acceleration of car,a_{c}=2.7 m/s^{2}

Let they meets at distance x from starting point.

So

For truck

x=(1/2)a_{T}(t-6)^{2}
............................(1)

and for car

x=(1/2)a_{c}(t)^{2}
.............................(2)

From (1) and (2)

(1/2)a_{T}(t-6)^{2}
=(1/2)a_{c}(t)^{2}

or

a_{T}(t-6)^{2} =a_{c}(t)^{2}

using a_{T}= 1 m/s^{2} and a_{c}=2.7
m/s^{2}

1.7t^{2} +12 t -36=0

on solving

t=2.27 sec

X=(1/2)a_{c}(t)^{2}=(1/2)2.72.27^{2}=6.96
m

Part a.

**1. 6.96 m**

**2. 2.27 sec**

Part b.

Velocity of truck when they meet is

V_{T}=(2a_{T}X)^{1/2}=(216.96)^{1/2}=**3.73
m/s**

Velocity of car when they meet is

V_{c}=(2a_{c}X)^{1/2}=(22.76.96)^{1/2}=
**6.13 m/s**

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