Question

A truck starts from rest. It travels a distance of 400m in 20s . Find its acceleration . Find the force acting on it if its mass is 7000

Answer 1

$\large\red{\underline{\underline{\rm{\green{Given}}}}}$

• Initial velocity (u) = 0 m/s. --(At rest)
• Distance travelled (s) = 400m
• Time taken (t) = 20s
• Mass (m) = 7000 kg

$\large\red{\underline{\underline{\rm{\green{To\ Find}}}}}$

• Acceleration (a)
• Force (F)

$\large\red{\underline{\underline{\rm{\green{Solution}}}}}$

Now,

On using 2nd equation of motion, we get,

➦ s = ut +1/2 at²

➭ 400 = 0×20 + 1/2a × (20)²

➭ 400 = a/2 × 400

➭ 400 = 400a/2

➭ 400 = 200a

➭ a = 400/200

➭ a = 2m/s²

Hence,

• Acceleration (a) = 2 m/s²

Now,

We know that,

➦ Force = Mass × Acceleration

➦ F = m × a

On putting above values in it, we get,

➭ F = 7000 × 2

➭ F = 14000 N

Hence,

• Force(F) = 14000 Newtons

Answer 2

$\Large{\underline{\underline{\tt{\red{Given}}}}}$

• Initial velocity (u) = 0
• Distance travelled (s) = 400m
• Time taken (t) = 20s
• Mass of truck (m) = 7000

$\Large{\underline{\underline{\tt{\red{Find\:out}}}}}$

• Find the acceleration
• Find the force acting on truck

$\Large{\underline{\underline{\tt{\red{Solution}}}}}$

According to the 2nd equation of motion

$\implies\sf s=ut+\dfrac{1}{2}at^2 \\ \\ \\ \implies\sf 400=0\times{20}+\dfrac{1}{2}\times{a}\times{(20)^2} \\ \\ \\ \implies\sf 400=\dfrac{1}{2}\times{400a} \\ \\ \\ \implies\sf 400 =200a \\ \\ \\ \implies\sf a=\cancel\dfrac{400}{200} \\ \\ \\ \implies\sf a=2m/s^2$

Now,

➞ Force = mass × acceleration

➞ Force = 7000 × 2

➞ Force = 14000N

Therefore ,

${\boxed{\bf{\purple{Acceleration\:of\:truck=2m/s^2}}}}$

${\boxed{\bf{\purple{Force\:acting\:on\:truck=14000N}}}}$