Question

if a truck starts from rest and it has acceleration of 4 m/s2 for 5 seconds, calculate its final velocity. What will be the distance travelled by the truck during the time interval?

Answer 1

Explanation:

u=0

a=4m/s2

t=5s

s=ut+1/2(at2)

0+1/2(4*25)

s=2*25

=50m

Answer 2

Given:

• Acceleration of the truck is 4m/s2
• Time Taken by the truck is 5 seconds

To Find:

• Final Velocity of truck
• Distance travelled by the truck during time interval

★ Formula Used :

$\nrightarrow\sf{v=u+at}$

$\nrightarrow\sf{{v}^{2}-{u}^{2}=2as}$

• v is Final velocity
• u is Initial velocity
• a is Acceleration
• t is Time taken
• s is Distance travelled

Solution:

As, the truck starts from rest; so it's initial speed will be 0m/s.

• Let's first find out the Final velocity of the truck.

$\: \: \: \: \: \longrightarrow\sf{v=u+at}$

$\: \: \: \: \: \longrightarrow\sf{v=0+4\times 5}$

$\: \: \: \: \: \longrightarrow\sf{v=4\times 5}$

$\: \: \: \: \: \longrightarrow\bold{v=20\: m/s}$

• Henceforth, Final Velocity of the truck is 20m/s

• Now, let's find the distance travelled by the truck.

$\: \: \: \: \: \implies\sf{{v}^{2}-{u}^{2}=2as}$

$\: \: \: \: \: \implies\sf{{20}^{2}-{0}^{2}=2\times 4\times s}$

$\: \: \: \: \: \implies\sf{400-0=8\times s}$

$\: \: \: \: \: \implies\sf{400=8\times s}$

$\: \: \: \: \: \implies\sf{\dfrac{400}{8}=s}$

$\: \: \: \: \: \implies\bold{s=50\: m}$

• Hence, Distance travelled by the truck is 50m.